Find last 3 digits of $ 2032^2031^2030^dots^2^1$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)mod Distributive Law, factoring $!!bmod!!:$ $ abbmod ac = a(bbmod c)$A power-exponential congruence equationHow to calculate $a^b$ (mod 10, 100 etc. ), where $a$ is not co-prime to $10$.Why is $x^100 = 1 mod 1000$ if $x < 1000$ and $gcd (x,1000) = 1$?Last 3 digits of Marsenne numbersShow that $a^varphi(n)+1equiv a,mod n$.Showing that $a^5$ and $a$ have the same last digit using Euler's Theorem.Find the last ten digits of this exponential tower.Euler theorem, finding last digitsRSA decryption coefficientFinding the last 4 digits of a huge power

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Find last 3 digits of $ 2032^2031^2030^dots^2^1$



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)mod Distributive Law, factoring $!!bmod!!:$ $ abbmod ac = a(bbmod c)$A power-exponential congruence equationHow to calculate $a^b$ (mod 10, 100 etc. ), where $a$ is not co-prime to $10$.Why is $x^100 = 1 mod 1000$ if $x < 1000$ and $gcd (x,1000) = 1$?Last 3 digits of Marsenne numbersShow that $a^varphi(n)+1equiv a,mod n$.Showing that $a^5$ and $a$ have the same last digit using Euler's Theorem.Find the last ten digits of this exponential tower.Euler theorem, finding last digitsRSA decryption coefficientFinding the last 4 digits of a huge power










7












$begingroup$


Find the last 3 digits of this number
$$
2032^2031^2030^dots^2^1
$$

So obviously we are looking for $x$ so that
$$
2032^2031^2030^dots^2^1 equiv x quad textmodhspace0.1cm 1000
$$

I also know that usually you use Euler' theorem here, but that only works when the numbers $a$ and $n$ are coprime, but $2032$ and $1000$ are not coprime? I can easily find $varphi(1000)$, that is not a problem. Am I looking for wrong numbers to be coprime here or is there another way instead of Euler' theorem?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
    $endgroup$
    – Mark Bennet
    Mar 24 at 21:20










  • $begingroup$
    @Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:59











  • $begingroup$
    Bah, that's no monster. Graham's Number is a monster!
    $endgroup$
    – Cort Ammon
    Mar 25 at 5:48















7












$begingroup$


Find the last 3 digits of this number
$$
2032^2031^2030^dots^2^1
$$

So obviously we are looking for $x$ so that
$$
2032^2031^2030^dots^2^1 equiv x quad textmodhspace0.1cm 1000
$$

I also know that usually you use Euler' theorem here, but that only works when the numbers $a$ and $n$ are coprime, but $2032$ and $1000$ are not coprime? I can easily find $varphi(1000)$, that is not a problem. Am I looking for wrong numbers to be coprime here or is there another way instead of Euler' theorem?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
    $endgroup$
    – Mark Bennet
    Mar 24 at 21:20










  • $begingroup$
    @Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:59











  • $begingroup$
    Bah, that's no monster. Graham's Number is a monster!
    $endgroup$
    – Cort Ammon
    Mar 25 at 5:48













7












7








7


1



$begingroup$


Find the last 3 digits of this number
$$
2032^2031^2030^dots^2^1
$$

So obviously we are looking for $x$ so that
$$
2032^2031^2030^dots^2^1 equiv x quad textmodhspace0.1cm 1000
$$

I also know that usually you use Euler' theorem here, but that only works when the numbers $a$ and $n$ are coprime, but $2032$ and $1000$ are not coprime? I can easily find $varphi(1000)$, that is not a problem. Am I looking for wrong numbers to be coprime here or is there another way instead of Euler' theorem?










share|cite|improve this question











$endgroup$




Find the last 3 digits of this number
$$
2032^2031^2030^dots^2^1
$$

So obviously we are looking for $x$ so that
$$
2032^2031^2030^dots^2^1 equiv x quad textmodhspace0.1cm 1000
$$

I also know that usually you use Euler' theorem here, but that only works when the numbers $a$ and $n$ are coprime, but $2032$ and $1000$ are not coprime? I can easily find $varphi(1000)$, that is not a problem. Am I looking for wrong numbers to be coprime here or is there another way instead of Euler' theorem?







number-theory totient-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 7:49









Asaf Karagila

308k33441775




308k33441775










asked Mar 24 at 20:59









Kristin PeterselKristin Petersel

363




363







  • 2




    $begingroup$
    The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
    $endgroup$
    – Mark Bennet
    Mar 24 at 21:20










  • $begingroup$
    @Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:59











  • $begingroup$
    Bah, that's no monster. Graham's Number is a monster!
    $endgroup$
    – Cort Ammon
    Mar 25 at 5:48












  • 2




    $begingroup$
    The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
    $endgroup$
    – Mark Bennet
    Mar 24 at 21:20










  • $begingroup$
    @Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:59











  • $begingroup$
    Bah, that's no monster. Graham's Number is a monster!
    $endgroup$
    – Cort Ammon
    Mar 25 at 5:48







2




2




$begingroup$
The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
$endgroup$
– Mark Bennet
Mar 24 at 21:20




$begingroup$
The common factor here is $8$ and you should be able show that he power is divisible by $8$, so the residue modulo $1000$ can be determined by looking at the residue modulo $1000/8=125$
$endgroup$
– Mark Bennet
Mar 24 at 21:20












$begingroup$
@Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
$endgroup$
– Bill Dubuque
Mar 24 at 23:59





$begingroup$
@Mark is correct, and in fact we can factor it out in a slick operational way that avoids using CRT by using the mod Distributive Law, as I show in my answer. This usually ends up being simpler than rotely applying CRT = Chinese Remainder when the base shares a common factor with the modulus.
$endgroup$
– Bill Dubuque
Mar 24 at 23:59













$begingroup$
Bah, that's no monster. Graham's Number is a monster!
$endgroup$
– Cort Ammon
Mar 25 at 5:48




$begingroup$
Bah, that's no monster. Graham's Number is a monster!
$endgroup$
– Cort Ammon
Mar 25 at 5:48










4 Answers
4






active

oldest

votes


















6












$begingroup$

$bmod 1000!: 32^large 2031^LARGE 2k!!equiv, 8left[dfraccolor#0a032^large 2031^LARGE 2k8 bmod color#0a0125right]! equiv 8left[dfraccolor#c00328bmod 125right]! equiv 32, $ by



$ ,beginalign !bmod color#0a0125!: color#0a032^large 2031^LARGE 2k!!
&equiv, 2^large 5cdot 2031^LARGE 2k! bmod 100 rm by 100 = phi(125) rm [Euler totient]\
&equiv,2^large 5(color#b6f2031^LARGE color#d4f2k! bmod 20) rm by mod Distributive Law\
&equiv,2^large 5(color#b6f1^LARGE k)equiv, color#c0032 rm by color#b6f2031^large 2!equiv 11^large 2equivcolor#b6f 1!!!pmod!20\
endalign $






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
    $endgroup$
    – Bill Dubuque
    Mar 24 at 22:50











  • $begingroup$
    How would you prove this?
    $endgroup$
    – Markus Punnar
    Mar 25 at 11:23










  • $begingroup$
    @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
    $endgroup$
    – Bill Dubuque
    Mar 25 at 13:05


















4












$begingroup$

It's a lot simpler than it looks. I shall call the number $N$.



You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.



$N$ is obviously a multiple of $8$, thus $Nequiv 0bmod 8$. Which leaves $bmod 125$.



The base $2032equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form



$2031^10k=(2030+1)^10k=(textbinomial expansion)=100m+1$



So $Nequiv 32^1equiv 32bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $Nequiv 32bmod 1000$. Meaning the last three digits were there all along, the $colorblue032$ in the base $2032$!






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    (+1) same answer I got, by means similar enough that I won't add another post here.
    $endgroup$
    – robjohn
    Mar 24 at 22:25











  • $begingroup$
    Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:27



















4












$begingroup$

Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.



$2032^monster$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.



$2032^monster = 0 pmod 8$ and so we just need to solve $2032^monster pmod 125$ and for that we can use Euler Theorem.



$phi(125=5^3) = (5-1)*5^3-1 = 100$.



So $2032^monster equiv 32^monster % 100$.



And $monster = 2031^littlemonsterequiv 31^littlemonsterpmod 100$



$31$ and $100$ are relatively prime and $phi(100)= 40$ so



$31^littlemonster equiv 31^littlemonster % 40 pmod 100$.



$littlemonster = 2030^smallmonster$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^smallmonster$ and $2^smallmonster|2030^smallmonster$.



So $littlemonster equiv 0 pmod 40$.



$2031^littlemonster equiv 31^0 equiv 1 pmod 100$



So $2032^monster equiv 32 pmod 125$



So $2032^monster equiv 0 pmod 8$ and $2032^monster equiv 32 pmod 125$.



As $8|32$ we are done. $2032^monster equiv 32 pmod 1000$.



and the last three digits are $032$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
    $endgroup$
    – Oscar Lanzi
    Mar 25 at 1:30










  • $begingroup$
    oops..............
    $endgroup$
    – fleablood
    Mar 25 at 11:58






  • 1




    $begingroup$
    Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
    $endgroup$
    – fleablood
    Mar 25 at 12:02


















2












$begingroup$

By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$mod 8$ and$mod 125$. Obviously



$$z_0:=2032^2031^2030^dots^2^1 equiv 0 pmod 8$$



What remains to be found is $x_0 in [0,124]$ in



$$z_0 equiv x_0 pmod 125.$$



As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With



$$z_1:=2031^2030^dots^2^1$$



and



$$phi(125)=100$$ the new problem becomes to find $x_1 in [0,99]$ in



$$z_1 equiv x_1 pmod 100$$



and then use



$$32^x_1 equiv x_0 pmod 125$$



to find $x_0$.



So this reduced the original problem$mod 1000$ to a smaller problem$mod 100$.



Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $mod 2$).



Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_i-1$, just as outlined for $x_1,x_0$ above.






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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    $bmod 1000!: 32^large 2031^LARGE 2k!!equiv, 8left[dfraccolor#0a032^large 2031^LARGE 2k8 bmod color#0a0125right]! equiv 8left[dfraccolor#c00328bmod 125right]! equiv 32, $ by



    $ ,beginalign !bmod color#0a0125!: color#0a032^large 2031^LARGE 2k!!
    &equiv, 2^large 5cdot 2031^LARGE 2k! bmod 100 rm by 100 = phi(125) rm [Euler totient]\
    &equiv,2^large 5(color#b6f2031^LARGE color#d4f2k! bmod 20) rm by mod Distributive Law\
    &equiv,2^large 5(color#b6f1^LARGE k)equiv, color#c0032 rm by color#b6f2031^large 2!equiv 11^large 2equivcolor#b6f 1!!!pmod!20\
    endalign $






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
      $endgroup$
      – Bill Dubuque
      Mar 24 at 22:50











    • $begingroup$
      How would you prove this?
      $endgroup$
      – Markus Punnar
      Mar 25 at 11:23










    • $begingroup$
      @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
      $endgroup$
      – Bill Dubuque
      Mar 25 at 13:05















    6












    $begingroup$

    $bmod 1000!: 32^large 2031^LARGE 2k!!equiv, 8left[dfraccolor#0a032^large 2031^LARGE 2k8 bmod color#0a0125right]! equiv 8left[dfraccolor#c00328bmod 125right]! equiv 32, $ by



    $ ,beginalign !bmod color#0a0125!: color#0a032^large 2031^LARGE 2k!!
    &equiv, 2^large 5cdot 2031^LARGE 2k! bmod 100 rm by 100 = phi(125) rm [Euler totient]\
    &equiv,2^large 5(color#b6f2031^LARGE color#d4f2k! bmod 20) rm by mod Distributive Law\
    &equiv,2^large 5(color#b6f1^LARGE k)equiv, color#c0032 rm by color#b6f2031^large 2!equiv 11^large 2equivcolor#b6f 1!!!pmod!20\
    endalign $






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
      $endgroup$
      – Bill Dubuque
      Mar 24 at 22:50











    • $begingroup$
      How would you prove this?
      $endgroup$
      – Markus Punnar
      Mar 25 at 11:23










    • $begingroup$
      @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
      $endgroup$
      – Bill Dubuque
      Mar 25 at 13:05













    6












    6








    6





    $begingroup$

    $bmod 1000!: 32^large 2031^LARGE 2k!!equiv, 8left[dfraccolor#0a032^large 2031^LARGE 2k8 bmod color#0a0125right]! equiv 8left[dfraccolor#c00328bmod 125right]! equiv 32, $ by



    $ ,beginalign !bmod color#0a0125!: color#0a032^large 2031^LARGE 2k!!
    &equiv, 2^large 5cdot 2031^LARGE 2k! bmod 100 rm by 100 = phi(125) rm [Euler totient]\
    &equiv,2^large 5(color#b6f2031^LARGE color#d4f2k! bmod 20) rm by mod Distributive Law\
    &equiv,2^large 5(color#b6f1^LARGE k)equiv, color#c0032 rm by color#b6f2031^large 2!equiv 11^large 2equivcolor#b6f 1!!!pmod!20\
    endalign $






    share|cite|improve this answer











    $endgroup$



    $bmod 1000!: 32^large 2031^LARGE 2k!!equiv, 8left[dfraccolor#0a032^large 2031^LARGE 2k8 bmod color#0a0125right]! equiv 8left[dfraccolor#c00328bmod 125right]! equiv 32, $ by



    $ ,beginalign !bmod color#0a0125!: color#0a032^large 2031^LARGE 2k!!
    &equiv, 2^large 5cdot 2031^LARGE 2k! bmod 100 rm by 100 = phi(125) rm [Euler totient]\
    &equiv,2^large 5(color#b6f2031^LARGE color#d4f2k! bmod 20) rm by mod Distributive Law\
    &equiv,2^large 5(color#b6f1^LARGE k)equiv, color#c0032 rm by color#b6f2031^large 2!equiv 11^large 2equivcolor#b6f 1!!!pmod!20\
    endalign $







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 24 at 23:40

























    answered Mar 24 at 22:48









    Bill DubuqueBill Dubuque

    214k29197658




    214k29197658







    • 2




      $begingroup$
      We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
      $endgroup$
      – Bill Dubuque
      Mar 24 at 22:50











    • $begingroup$
      How would you prove this?
      $endgroup$
      – Markus Punnar
      Mar 25 at 11:23










    • $begingroup$
      @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
      $endgroup$
      – Bill Dubuque
      Mar 25 at 13:05












    • 2




      $begingroup$
      We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
      $endgroup$
      – Bill Dubuque
      Mar 24 at 22:50











    • $begingroup$
      How would you prove this?
      $endgroup$
      – Markus Punnar
      Mar 25 at 11:23










    • $begingroup$
      @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
      $endgroup$
      – Bill Dubuque
      Mar 25 at 13:05







    2




    2




    $begingroup$
    We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
    $endgroup$
    – Bill Dubuque
    Mar 24 at 22:50





    $begingroup$
    We used twice: $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law $ $
    $endgroup$
    – Bill Dubuque
    Mar 24 at 22:50













    $begingroup$
    How would you prove this?
    $endgroup$
    – Markus Punnar
    Mar 25 at 11:23




    $begingroup$
    How would you prove this?
    $endgroup$
    – Markus Punnar
    Mar 25 at 11:23












    $begingroup$
    @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
    $endgroup$
    – Bill Dubuque
    Mar 25 at 13:05




    $begingroup$
    @Markus Prove what? If you mean the mod Distributive Law then follow the link in my prior comment.
    $endgroup$
    – Bill Dubuque
    Mar 25 at 13:05











    4












    $begingroup$

    It's a lot simpler than it looks. I shall call the number $N$.



    You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.



    $N$ is obviously a multiple of $8$, thus $Nequiv 0bmod 8$. Which leaves $bmod 125$.



    The base $2032equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form



    $2031^10k=(2030+1)^10k=(textbinomial expansion)=100m+1$



    So $Nequiv 32^1equiv 32bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $Nequiv 32bmod 1000$. Meaning the last three digits were there all along, the $colorblue032$ in the base $2032$!






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      (+1) same answer I got, by means similar enough that I won't add another post here.
      $endgroup$
      – robjohn
      Mar 24 at 22:25











    • $begingroup$
      Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
      $endgroup$
      – Bill Dubuque
      Mar 24 at 23:27
















    4












    $begingroup$

    It's a lot simpler than it looks. I shall call the number $N$.



    You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.



    $N$ is obviously a multiple of $8$, thus $Nequiv 0bmod 8$. Which leaves $bmod 125$.



    The base $2032equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form



    $2031^10k=(2030+1)^10k=(textbinomial expansion)=100m+1$



    So $Nequiv 32^1equiv 32bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $Nequiv 32bmod 1000$. Meaning the last three digits were there all along, the $colorblue032$ in the base $2032$!






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      (+1) same answer I got, by means similar enough that I won't add another post here.
      $endgroup$
      – robjohn
      Mar 24 at 22:25











    • $begingroup$
      Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
      $endgroup$
      – Bill Dubuque
      Mar 24 at 23:27














    4












    4








    4





    $begingroup$

    It's a lot simpler than it looks. I shall call the number $N$.



    You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.



    $N$ is obviously a multiple of $8$, thus $Nequiv 0bmod 8$. Which leaves $bmod 125$.



    The base $2032equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form



    $2031^10k=(2030+1)^10k=(textbinomial expansion)=100m+1$



    So $Nequiv 32^1equiv 32bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $Nequiv 32bmod 1000$. Meaning the last three digits were there all along, the $colorblue032$ in the base $2032$!






    share|cite|improve this answer











    $endgroup$



    It's a lot simpler than it looks. I shall call the number $N$.



    You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.



    $N$ is obviously a multiple of $8$, thus $Nequiv 0bmod 8$. Which leaves $bmod 125$.



    The base $2032equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form



    $2031^10k=(2030+1)^10k=(textbinomial expansion)=100m+1$



    So $Nequiv 32^1equiv 32bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $Nequiv 32bmod 1000$. Meaning the last three digits were there all along, the $colorblue032$ in the base $2032$!







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 24 at 21:36

























    answered Mar 24 at 21:30









    Oscar LanziOscar Lanzi

    13.6k12136




    13.6k12136







    • 2




      $begingroup$
      (+1) same answer I got, by means similar enough that I won't add another post here.
      $endgroup$
      – robjohn
      Mar 24 at 22:25











    • $begingroup$
      Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
      $endgroup$
      – Bill Dubuque
      Mar 24 at 23:27













    • 2




      $begingroup$
      (+1) same answer I got, by means similar enough that I won't add another post here.
      $endgroup$
      – robjohn
      Mar 24 at 22:25











    • $begingroup$
      Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
      $endgroup$
      – Bill Dubuque
      Mar 24 at 23:27








    2




    2




    $begingroup$
    (+1) same answer I got, by means similar enough that I won't add another post here.
    $endgroup$
    – robjohn
    Mar 24 at 22:25





    $begingroup$
    (+1) same answer I got, by means similar enough that I won't add another post here.
    $endgroup$
    – robjohn
    Mar 24 at 22:25













    $begingroup$
    Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:27





    $begingroup$
    Worth emphasis is that arguments like this can be presented completely operationally by employing the mod Distributive Law, and this clarifies and simplifies the arithmetic - see my answer.
    $endgroup$
    – Bill Dubuque
    Mar 24 at 23:27












    4












    $begingroup$

    Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.



    $2032^monster$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.



    $2032^monster = 0 pmod 8$ and so we just need to solve $2032^monster pmod 125$ and for that we can use Euler Theorem.



    $phi(125=5^3) = (5-1)*5^3-1 = 100$.



    So $2032^monster equiv 32^monster % 100$.



    And $monster = 2031^littlemonsterequiv 31^littlemonsterpmod 100$



    $31$ and $100$ are relatively prime and $phi(100)= 40$ so



    $31^littlemonster equiv 31^littlemonster % 40 pmod 100$.



    $littlemonster = 2030^smallmonster$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^smallmonster$ and $2^smallmonster|2030^smallmonster$.



    So $littlemonster equiv 0 pmod 40$.



    $2031^littlemonster equiv 31^0 equiv 1 pmod 100$



    So $2032^monster equiv 32 pmod 125$



    So $2032^monster equiv 0 pmod 8$ and $2032^monster equiv 32 pmod 125$.



    As $8|32$ we are done. $2032^monster equiv 32 pmod 1000$.



    and the last three digits are $032$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
      $endgroup$
      – Oscar Lanzi
      Mar 25 at 1:30










    • $begingroup$
      oops..............
      $endgroup$
      – fleablood
      Mar 25 at 11:58






    • 1




      $begingroup$
      Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
      $endgroup$
      – fleablood
      Mar 25 at 12:02















    4












    $begingroup$

    Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.



    $2032^monster$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.



    $2032^monster = 0 pmod 8$ and so we just need to solve $2032^monster pmod 125$ and for that we can use Euler Theorem.



    $phi(125=5^3) = (5-1)*5^3-1 = 100$.



    So $2032^monster equiv 32^monster % 100$.



    And $monster = 2031^littlemonsterequiv 31^littlemonsterpmod 100$



    $31$ and $100$ are relatively prime and $phi(100)= 40$ so



    $31^littlemonster equiv 31^littlemonster % 40 pmod 100$.



    $littlemonster = 2030^smallmonster$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^smallmonster$ and $2^smallmonster|2030^smallmonster$.



    So $littlemonster equiv 0 pmod 40$.



    $2031^littlemonster equiv 31^0 equiv 1 pmod 100$



    So $2032^monster equiv 32 pmod 125$



    So $2032^monster equiv 0 pmod 8$ and $2032^monster equiv 32 pmod 125$.



    As $8|32$ we are done. $2032^monster equiv 32 pmod 1000$.



    and the last three digits are $032$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
      $endgroup$
      – Oscar Lanzi
      Mar 25 at 1:30










    • $begingroup$
      oops..............
      $endgroup$
      – fleablood
      Mar 25 at 11:58






    • 1




      $begingroup$
      Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
      $endgroup$
      – fleablood
      Mar 25 at 12:02













    4












    4








    4





    $begingroup$

    Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.



    $2032^monster$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.



    $2032^monster = 0 pmod 8$ and so we just need to solve $2032^monster pmod 125$ and for that we can use Euler Theorem.



    $phi(125=5^3) = (5-1)*5^3-1 = 100$.



    So $2032^monster equiv 32^monster % 100$.



    And $monster = 2031^littlemonsterequiv 31^littlemonsterpmod 100$



    $31$ and $100$ are relatively prime and $phi(100)= 40$ so



    $31^littlemonster equiv 31^littlemonster % 40 pmod 100$.



    $littlemonster = 2030^smallmonster$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^smallmonster$ and $2^smallmonster|2030^smallmonster$.



    So $littlemonster equiv 0 pmod 40$.



    $2031^littlemonster equiv 31^0 equiv 1 pmod 100$



    So $2032^monster equiv 32 pmod 125$



    So $2032^monster equiv 0 pmod 8$ and $2032^monster equiv 32 pmod 125$.



    As $8|32$ we are done. $2032^monster equiv 32 pmod 1000$.



    and the last three digits are $032$.






    share|cite|improve this answer











    $endgroup$



    Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.



    $2032^monster$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.



    $2032^monster = 0 pmod 8$ and so we just need to solve $2032^monster pmod 125$ and for that we can use Euler Theorem.



    $phi(125=5^3) = (5-1)*5^3-1 = 100$.



    So $2032^monster equiv 32^monster % 100$.



    And $monster = 2031^littlemonsterequiv 31^littlemonsterpmod 100$



    $31$ and $100$ are relatively prime and $phi(100)= 40$ so



    $31^littlemonster equiv 31^littlemonster % 40 pmod 100$.



    $littlemonster = 2030^smallmonster$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^smallmonster$ and $2^smallmonster|2030^smallmonster$.



    So $littlemonster equiv 0 pmod 40$.



    $2031^littlemonster equiv 31^0 equiv 1 pmod 100$



    So $2032^monster equiv 32 pmod 125$



    So $2032^monster equiv 0 pmod 8$ and $2032^monster equiv 32 pmod 125$.



    As $8|32$ we are done. $2032^monster equiv 32 pmod 1000$.



    and the last three digits are $032$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 25 at 12:00

























    answered Mar 24 at 22:56









    fleabloodfleablood

    1




    1







    • 1




      $begingroup$
      Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
      $endgroup$
      – Oscar Lanzi
      Mar 25 at 1:30










    • $begingroup$
      oops..............
      $endgroup$
      – fleablood
      Mar 25 at 11:58






    • 1




      $begingroup$
      Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
      $endgroup$
      – fleablood
      Mar 25 at 12:02












    • 1




      $begingroup$
      Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
      $endgroup$
      – Oscar Lanzi
      Mar 25 at 1:30










    • $begingroup$
      oops..............
      $endgroup$
      – fleablood
      Mar 25 at 11:58






    • 1




      $begingroup$
      Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
      $endgroup$
      – fleablood
      Mar 25 at 12:02







    1




    1




    $begingroup$
    Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
    $endgroup$
    – Oscar Lanzi
    Mar 25 at 1:30




    $begingroup$
    Good idea, but you rendered $2031^2030^··· bmod 125$. You need $2032^2031^2030^··· bmod 125$.
    $endgroup$
    – Oscar Lanzi
    Mar 25 at 1:30












    $begingroup$
    oops..............
    $endgroup$
    – fleablood
    Mar 25 at 11:58




    $begingroup$
    oops..............
    $endgroup$
    – fleablood
    Mar 25 at 11:58




    1




    1




    $begingroup$
    Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
    $endgroup$
    – fleablood
    Mar 25 at 12:02




    $begingroup$
    Actually I rendered it as $2032^2031^...equiv 31^2031^...pmod125$... which is still an error.
    $endgroup$
    – fleablood
    Mar 25 at 12:02











    2












    $begingroup$

    By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$mod 8$ and$mod 125$. Obviously



    $$z_0:=2032^2031^2030^dots^2^1 equiv 0 pmod 8$$



    What remains to be found is $x_0 in [0,124]$ in



    $$z_0 equiv x_0 pmod 125.$$



    As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With



    $$z_1:=2031^2030^dots^2^1$$



    and



    $$phi(125)=100$$ the new problem becomes to find $x_1 in [0,99]$ in



    $$z_1 equiv x_1 pmod 100$$



    and then use



    $$32^x_1 equiv x_0 pmod 125$$



    to find $x_0$.



    So this reduced the original problem$mod 1000$ to a smaller problem$mod 100$.



    Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $mod 2$).



    Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_i-1$, just as outlined for $x_1,x_0$ above.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$mod 8$ and$mod 125$. Obviously



      $$z_0:=2032^2031^2030^dots^2^1 equiv 0 pmod 8$$



      What remains to be found is $x_0 in [0,124]$ in



      $$z_0 equiv x_0 pmod 125.$$



      As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With



      $$z_1:=2031^2030^dots^2^1$$



      and



      $$phi(125)=100$$ the new problem becomes to find $x_1 in [0,99]$ in



      $$z_1 equiv x_1 pmod 100$$



      and then use



      $$32^x_1 equiv x_0 pmod 125$$



      to find $x_0$.



      So this reduced the original problem$mod 1000$ to a smaller problem$mod 100$.



      Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $mod 2$).



      Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_i-1$, just as outlined for $x_1,x_0$ above.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$mod 8$ and$mod 125$. Obviously



        $$z_0:=2032^2031^2030^dots^2^1 equiv 0 pmod 8$$



        What remains to be found is $x_0 in [0,124]$ in



        $$z_0 equiv x_0 pmod 125.$$



        As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With



        $$z_1:=2031^2030^dots^2^1$$



        and



        $$phi(125)=100$$ the new problem becomes to find $x_1 in [0,99]$ in



        $$z_1 equiv x_1 pmod 100$$



        and then use



        $$32^x_1 equiv x_0 pmod 125$$



        to find $x_0$.



        So this reduced the original problem$mod 1000$ to a smaller problem$mod 100$.



        Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $mod 2$).



        Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_i-1$, just as outlined for $x_1,x_0$ above.






        share|cite|improve this answer









        $endgroup$



        By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$mod 8$ and$mod 125$. Obviously



        $$z_0:=2032^2031^2030^dots^2^1 equiv 0 pmod 8$$



        What remains to be found is $x_0 in [0,124]$ in



        $$z_0 equiv x_0 pmod 125.$$



        As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With



        $$z_1:=2031^2030^dots^2^1$$



        and



        $$phi(125)=100$$ the new problem becomes to find $x_1 in [0,99]$ in



        $$z_1 equiv x_1 pmod 100$$



        and then use



        $$32^x_1 equiv x_0 pmod 125$$



        to find $x_0$.



        So this reduced the original problem$mod 1000$ to a smaller problem$mod 100$.



        Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $mod 2$).



        Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_i-1$, just as outlined for $x_1,x_0$ above.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 24 at 21:33









        IngixIngix

        5,232259




        5,232259



























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