Problem 1
(2013 PUMaC, Number Theory Problem 2) Find the smallest \(n\) such that \(2013^n\) ends in \(001\).
Solution 1
It is easy to prove that the problem is equivalent to finding the smallest \(n>0\) such that \(2013^n\equiv1\pmod{1000}\).
\[\begin{align}
2013^n\equiv1\pmod{1000}&\Longleftrightarrow2013^n\equiv1\pmod{2^3}\land2013^n\equiv1\pmod{5^3}\\
&\Longleftrightarrow5^n\equiv1\pmod{2^3}\land13^n\equiv1\pmod{5^3}\\
&\Longleftrightarrow2^3\mid5^n-1^n\land5^3\mid13^n-1^n\\
&\Longleftrightarrow\nu_2(5^n-1^n)\ge3\land\nu_5(13^n-1^n)\ge3\\
\end{align}
\]
If \(n\equiv0\pmod2\), applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_2(5^n-1^n)&=\nu_2(4)+\nu_2(6)+\nu_2(n)-1\\
&=2+1+\nu_2(n)-1\\
&\ge3
\end{align}
\]
If \(n\equiv1\pmod2\), applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_2(5^n-1^n)&=\nu_2(4)\\
&=2\\
&<3
\end{align}
\]
Therefore,
\[\nu_2(5^n-1^n)\ge3\Longleftrightarrow2\mid n
\]
\[\begin{align}
\nu_5(13^n-1^n)\ge3&\Longrightarrow3^n\equiv1\pmod5\\
&\Longleftrightarrow\delta_5(3)\mid n\\
&\Longleftrightarrow4\mid n
\end{align}
\]
Therefore,
\[\begin{align}
\nu_5(13^n-1^n)\ge3&\Longleftrightarrow4\mid n\land\nu_5(13^n-1^n)\ge3\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land\nu_5(13^n-1^n)\ge3\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land\nu_5((13^4)^k-(1^4)^k)\ge3\\
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_5((13^4)^k-(1^4)^k)&=\nu_5(13^4-1^4)+\nu_5(k)\\
&=\nu_5(13-1)+\nu_5(13+1)+\nu_5(13^2+1^2)+\nu_5(k)\\
&=1+\nu_5(k)
\end{align}
\]
Therefore,
\[\begin{align}
\nu_5(13^n-1^n)\ge3&\Longleftrightarrow\exist k\in\Z,n=4k\land1+\nu_5(k)\ge3\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land25\mid k\\
&\Longleftrightarrow100\mid n
\end{align}
\]
Therefore,
\[\begin{align}
2013^n\equiv1\pmod{1000}&\Longleftrightarrow2\mid n\land100\mid n\\
&\Longleftrightarrow100\mid n
\end{align}
\]
The smallest \(n>0\) such that \(100\mid n\) is \(100\).
Problem 2
(2020 AIME I, Problem 12) Let \(n\) be the least positive integer such that \(149^n-2^n\) is divisible by \(3^35^57^7\). Find the number of positive divisors of \(n\).
Solution 2
\[\begin{align}
3^35^57^7\mid(149^n-2^n)&\Longleftrightarrow3^3\mid149^n-2^n\land5^5\mid149^n-2^n\land7^7\mid149^n-2^n\\
&\Longleftrightarrow\nu_3(149^n-2^n)\ge3\land\nu_5(149^n-2^n)\ge5\land\nu_7(149^n-2^n)\ge7
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_3(149^n-2^n)&=\nu_3(147)+\nu_3(n)\\
&=1+\nu_3(n)
\end{align}
\]
Therefore,
\[\begin{align}
\nu_3(149^n-2^n)\ge3&\Longleftrightarrow1+\nu_3(n)\ge3\\
&\Longleftrightarrow3^2\mid n
\end{align}
\]
\[\nu_5(149^n-2^n)\ge5\Longrightarrow149^n\equiv2^n\pmod5
\]
Since
\[149^n\equiv\begin{cases}
1\pmod5, &n\equiv0\pmod2\\
4\pmod5, &n\equiv1\pmod2
\end{cases}
\]
\[2^n\equiv\begin{cases}
1\pmod5, &n\equiv0\pmod4\\
2\pmod5, &n\equiv1\pmod4\\
4\pmod5, &n\equiv2\pmod4\\
3\pmod5, &n\equiv3\pmod4
\end{cases}
\]
it follows that
\[149^n\equiv2^n\pmod5\Longleftrightarrow4\mid n
\]
Therefore,
\[\begin{align}
\nu_5(149^n-2^n)\ge5&\Longleftrightarrow4\mid n\land\nu_5(149^n-2^n)\ge5\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land\nu_5(149^n-2^n)\ge5\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land\nu_5((149^4)^n-(2^4)^n)\ge5\\
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_5((149^4)^k-(2^4)^k)&=\nu_5(149^4-2^4)+\nu_5(k)\\
&=\nu_5(149-2)+\nu_5(149+2)+\nu_5(149^2+2^2)+\nu_5(k)\\
&=1+\nu_5(k)
\end{align}
\]
Therefore,
\[\begin{align}
\nu_5(149^n-2^n)\ge5&\Longleftrightarrow\exist k\in\Z,n=4k\land1+\nu_5(k)\ge5\\
&\Longleftrightarrow\exist k\in\Z,n=4k\land5^4\mid k\\
&\Longleftrightarrow4\cdot5^4\mid n\\
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_7(149^n-2^n)&=\nu_7(147)+\nu_7(n)\\
&=2+\nu_7(n)
\end{align}
\]
Therefore,
\[\begin{align}
\nu_7(149^n-2^n)\ge7&\Longleftrightarrow2+\nu_7(n)\ge7\\
&\Longleftrightarrow7^5\mid n
\end{align}
\]
Therefore,
\[\begin{align}
3^35^57^7\mid149^n-2^n&\Longleftrightarrow3^2\mid n\land4\cdot5^4\mid n\land 7^5\mid n\\
&\Longleftrightarrow2^23^25^47^5\mid n
\end{align}
\]
The smallest \(n>0\) such that \(2^23^25^47^5\mid n\) is \(2^23^25^47^5\), it has \((2+1)(2+1)(4+1)(5+1)=270\) divisors.
Problem 3
(1991 IMOSL) Find the largest \(k\) such that
\[1991^k\mid1990^{1991^{1992}}+1992^{1991^{1990}}
\]
Solution 3
\[\begin{align}
1991^k\mid1990^{1991^{1992}}+1992^{1991^{1990}}&\Longleftrightarrow11^k\mid1990^{1991^{1992}}+1992^{1991^{1990}}\land181^k\mid1990^{1991^{1992}}+1992^{1991^{1990}}\\
&\Longleftrightarrow\nu_{11}(1990^{1991^{1992}}+1992^{1991^{1990}})\ge k\land\nu_{181}(1990^{1991^{1992}}+1992^{1991^{1990}})\ge k\\
\end{align}
\]
\[\begin{align}
1990^{1991^{1992}}+1992^{1991^{1990}}&=1990^{1991^{1990}1991^2}+1992^{1991^{1990}}\\
&=(1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}}
\end{align}
\]
Therefore,
\[\nu_{11}(1990^{1991^{1992}}+1992^{1991^{1990}})=\nu_{11}((1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}})
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_{11}((1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}})&=\nu_{11}(1990^{1991^2}+1992)+1990\nu_{11}(1991)\\
&=\nu_{11}(\sum_{n=0}^{1991^2}\binom{1991^2}{n}1991^n(-1)^{1991^2-n}+1992)+1990\\
&=\nu_{11}(\sum_{n=1}^{1991^2}\binom{1991^2}{n}1991^n(-1)^{1991^2-n}+1991)+1990\\
&=\nu_{11}(1991)+\nu_{11}(\sum_{n=1}^{1991^2}\binom{1991^2}{n}1991^{n-1}(-1)^{1991^2-n}+1)+1990\\
&=1+0+1990\\
&=1991\\
\end{align}
\]
Similarly, it follows that
\[\nu_{181}(1990^{1991^{1992}}+1992^{1991^{1990}})=1991
\]
Therefore,
\[\begin{align}
1991^k\mid1990^{1991^{1992}}+1992^{1991^{1990}}&\Longleftrightarrow1991\ge k\land1991\ge k\\
&\Longleftrightarrow1991\ge k
\end{align}
\]
The largest \(k\) such that \(1991\ge k\) is \(1991\).
Problem 4
(2006 KMO first round, Problem 6) If \(m\) is an integer such that
\[3^m\mid7^{3^{527}}-1
\]
find the maximum possible value of \(m\).
Solution 4
\[3^m\mid7^{3^{527}}-1\Longleftrightarrow\nu_3(7^{3^{527}}-1^{3^{527}})\ge m
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_3(7^{3^{527}}-1^{3^{527}})&=\nu_3(6)+\nu_3(3^{527})\\
&=1+527\\
&=528\\
\end{align}
\]
Therefore,
\[3^m\mid7^{3^{527}}-1\Longleftrightarrow528\ge m
\]
The largest integer \(m\) such that \(528\ge m\) is \(528\).
Problem 5
Find the sum of all the divisors \(d\) of \(N=19^{88}-1\) which are of the form \(d=2^a3^b\) with \(a,b\in\N\).
Solution 5
\[\begin{align}
2^a3^b\mid19^{88}-1&\Longleftrightarrow2^a\mid19^{88}-1\land3^b\mid19^{88}-1\\
&\Longleftrightarrow\nu_2(19^{88}-1^{88})\ge a\land\nu_3(19^{88}-1^{88})\ge b
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_2(19^{88}-1^{88})&=\nu_2(19-1)+\nu_2(19+1)+\nu_2(88)-1\\
&=1+2+3-1\\
&=5
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_3(19^{88}-1^{88})&=\nu_3(19-1)+\nu_3(88)\\
&=2\\
\end{align}
\]
Therefore,
\[2^a3^b\mid19^{88}-1\Longleftrightarrow5\ge a\land2\ge b
\]
The sum of all the divisors \(d\) of the form \(d=2^a3^b\) is
\[\begin{align}
\sum_{a=0}^5\sum_{b=0}^22^a3^b&=(\sum_{a=0}^52^a)(\sum_{b=0}^23^b)\\
&=63\cdot13\\
&=819
\end{align}
\]
Problem 6
Let \(k\) be a positive integer. Find all positive integers \(n\) such that \(3^k\mid2^n-1\).
Solution 6
\[\begin{align}
3^k\mid2^n-1&\Longrightarrow3\mid2^n-1\\
&\Longleftrightarrow2^n\equiv1\pmod3\\
&\Longleftrightarrow\delta_3(2)\mid n\\
&\Longleftrightarrow2\mid n\\
\end{align}
\]
Therefore,
\[\begin{align}
3^k\mid2^n-1&\Longleftrightarrow2\mid n\land3^k\mid2^n-1\\
&\Longleftrightarrow\exist m\in\Z,n=2m\land3^k\mid2^n-1\\
&\Longleftrightarrow\exist m\in\Z,n=2m\land3^k\mid4^m-1\\
&\Longleftrightarrow\exist m\in\Z,n=2m\land\nu_3(4^m-1^m)\ge k\\
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_3(4^m-1^m)&=\nu_3(4-1)+\nu_3(m)\\
&=1+\nu_3(m)\\
\end{align}
\]
Therefore,
\[\begin{align}
3^k\mid(2^n-1)&\Longleftrightarrow\exist m\in\Z,n=2m\land1+\nu_3(m)\ge k\\
&\Longleftrightarrow\exist m\in\Z,n=2m\land3^{k-1}\mid m\\
&\Longleftrightarrow2\cdot3^{k-1}\mid m
\end{align}
\]
Problem 7
Find all primes \(p,q\) such that
\[\frac{(5^p-2^p)(5^q-2^q)}{pq}
\]
is an integer.
Solution 7
Without the loss of generality, let \(p\le q\).
\[\frac{(5^p-2^p)(5^q-2^q)}{pq}\in\Z\Longleftrightarrow pq\mid(5^p-2^p)(5^q-2^q)
\]
Since
\[(5^p-2^p)(5^q-2^q)\equiv1\pmod2
\]
it follows that
\[pq\mid(5^p-2^p)(5^q-2^q)\Longrightarrow p\ne2\land q\ne2
\]
\[\begin{align}
pq\mid(5^p-2^p)(5^q-2^q)&\Longrightarrow p\mid(5^p-2^p)(5^q-2^q)\\
&\Longleftrightarrow(5^p-2^p)(5^q-2^q)\equiv0\pmod p
\end{align}
\]
Applying Fermat's Little Theorem yields
\[5^p-2^p\equiv3\pmod p
\]
Therefore, if \(p\ne 3\),
\[\begin{align}
(5^p-2^p)(5^q-2^q)\equiv0\pmod p&\Longleftrightarrow5^q-2^q\equiv0\pmod p\\
&\Longleftrightarrow\left(\frac{5}{2}\right)^q\equiv1\pmod p\\
&\Longleftrightarrow\delta_p\left(\frac{5}{2}\right)\mid q\\
&\Longleftrightarrow\delta_p\left(\frac{5}{2}\right)=q\\
\end{align}
\]
Applying Fermat's Little Theorem yields
\[\left(\frac{5}{2}\right)^{p-1}\equiv1\pmod p
\]
\[\left(\frac{5}{2}\right)^{p-1}\equiv1\pmod p\Longleftrightarrow\delta_p\left(\frac{5}{2}\right)\mid p-1
\]
Therefore,
\[\begin{align}
(5^p-2^p)(5^q-2^q)\equiv0\pmod p&\Longrightarrow q\mid p-1\\
&\Longrightarrow q<p
\end{align}
\]
By contradiction, \(p=3\).
Therefore,
\[\begin{align}
pq\mid(5^p-2^p)(5^q-2^q)&\Longleftrightarrow39(5^q-2^q)\equiv0\pmod q\\
&\Longleftrightarrow q\mid117\\
&\Longleftrightarrow q=3\lor q=13\\
\end{align}
\]
Therefore, all primes \(p,q\) such that
\[\frac{(5^p-2^p)(5^q-2^q)}{pq}
\]
is an integer are \(\langle p,q\rangle=\langle3,3\rangle\), \(\langle p,q\rangle=\langle3,13\rangle\) and \(\langle p,q\rangle=\langle13,3\rangle\).
Problem 8
Find all positive integers \(n\) such that \(3^n-1\) is divisible by \(2^n\).
Solution 8
\[2^n\mid3^n-1\Longleftrightarrow\nu_2(3^n-1^n)\ge n
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_2(3^n-1^n)&=\begin{cases}
\nu_2(3-1)+\nu_2(3+1)+\nu_2(n)-1,&n\equiv0\pmod2\\
\nu_2(3-1),&n\equiv1\pmod2\\
\end{cases}\\
&=\begin{cases}
\nu_2(n)+2,&n\equiv0\pmod2\\
1,&n\equiv1\pmod2\\
\end{cases}
\end{align}
\]
Therefore,
\[\begin{align}
\nu_2(3^n-1^n)\ge n&\Longleftrightarrow(n\equiv0\pmod2\land\nu_2(n)+2\ge n)\lor(n\equiv1\pmod2\land1\ge n)\\
&\Longleftrightarrow(n\equiv0\pmod2\land 2^{n-2}\mid n)\lor n=1\\
&\Longleftrightarrow n=1\lor n=2\lor n=4
\end{align}
\]
Therefore,
\[2^n\mid3^n-1\Longleftrightarrow n=1\lor n=2\lor n=4
\]
Problem 9
Find all positive integers \(a\) such that \(\frac{5^a+1}{3^a}\) is a positive integer.
Solution 9
\[\begin{align}
\frac{5^a+1}{3^a}\in\Z_+&\Longleftrightarrow3^a\mid5^a+1\\
&\Longrightarrow5^a+1\equiv0\pmod3\\
&\Longleftrightarrow2^a\equiv2\pmod3\\
&\Longleftrightarrow a\equiv1\pmod2
\end{align}
\]
Therefore,
\[\begin{align}
3^a\mid5^a+1&\Longleftrightarrow a\equiv1\pmod2\land 3^a\mid5^a+1\\
&\Longleftrightarrow a\equiv1\pmod2\land\nu_3(5^a+1^a)\ge a\\
\end{align}
\]
Applying Lifting The Exponent Lemma yields that, if \(a\equiv1\pmod 2\),
\[\begin{align}
\nu_3(5^a+1^a)&=\nu_3(6)+\nu_3(a)\\
&=1+\nu_3(a)
\end{align}
\]
Therefore,
\[\begin{align}
3^a\mid5^a+1&\Longleftrightarrow1+\nu_3(a)\ge a\\
&\Longleftrightarrow3^{a-1}\mid a\\
&\Longleftrightarrow a=1
\end{align}
\]
Therefore, the only positive integer \(a\) such that \(\frac{5^a+1}{3^a}\) is a positive integer is \(a=1\).
Problem 10
(PUMaC 2023 Division A, Individual Finals Problem 1) Let \(p>3\) be a prime and \(k\ge0\) an integer. Find the multiplicity of \(X-1\) in the factorization of
\[f(X)=X^{3p^k-1}+X^{3p^k-2}+\dots+X+1
\]
modulo \(p\).
Solution 10
\[\begin{align}
f(X)&\equiv X^{3p^k-1}+X^{3p^k-2}+\dots+X+1\pmod p\\
&\equiv\frac{X^{3p^k}-1}{X-1}\pmod p\\
&\equiv\frac{(X^3)^{p^k}-1^{p^k}}{X-1}\pmod p\\
&\equiv\frac{(X^3-1)^{p^k}}{X-1}\pmod p\\
&\equiv(X-1)^{p^k-1}(X^2+X+1)^{p^k}\pmod p\\
\end{align}
\]
Since when \(X=1\), \(X-1\equiv0\pmod p\), \(X^2+X+1\not\equiv0\pmod p\), it follows that the multiplicity of \(X-1\) in the factorization of \(f(X)\) is \(p^k-1\), modulo \(p\).
Problem 11
(2017 CGMO, Problem 1) Find all positive integers \(n\) such that for every positive odd integer \(a\), we have
\[2^{2017}\mid a^n-1
\]
Solution 11
\[\forall a,2^{2017}\mid a^n-1\Longleftrightarrow\forall a,\nu_2(a^n-1^n)\ge2017
\]
Applying Lifting The Exponent Lemma yields
\[\nu_2(a^n-1^n)=\begin{cases}
\nu_2(a-1)+\nu_2(a+1)+\nu_2(n)-1,&n\equiv0\pmod2\\
\nu_2(a-1),&n\equiv1\pmod2
\end{cases}
\]
Therefore,
\[\begin{align}
\forall a,2^{2017}\mid a^n-1&\Longleftrightarrow(n\equiv0\pmod2\land\forall a,\nu_2(a-1)+\nu_2(a+1)+\nu_2(n)-1\ge2017)\lor(n\equiv1\pmod2\land\forall a,\nu_2(a-1)\ge2017)\\
&\Longleftrightarrow n\equiv0\pmod2\land\forall a,\nu_2(n)\ge2018-\nu_2(a-1)-\nu_2(a+1)\\
&\Longleftrightarrow n\equiv0\pmod2\land\nu_2(n)\ge2018-\min_a\{\nu_2(a-1)+\nu_2(a+1)\}\\
&\Longleftrightarrow n\equiv0\pmod2\land\nu_2(n)\ge2015\\
&\Longleftrightarrow2^{2015}\mid n
\end{align}
\]
Problem 12
(2010 China Western Mathematical Olympiad, Problem 1) Let \(m,k\ge0,p=2^{2^m}+1\) a prime. Prove:
- \(2^{2^{m+1}p^k}\equiv1\pmod{p^{k+1}}\);
- The smallest \(n\) such that \(2^n\equiv1\pmod{p^{k+1}}\) is \(2^{m+1}p^k\).
Solution 12
\[\begin{align}
2^{2^{m+1}p^k}\equiv1\pmod{p^{k+1}}&\Longleftrightarrow p^{k+1}\mid2^{2^{m+1}p^k}-1\\
&\Longleftrightarrow\nu_p((2^{2^{m+1}})^{p^k}-1^{p^k})\ge k+1
\end{align}
\]
Applying Lifting The Exponent Lemma yields
\[\begin{align}
\nu_p((2^{2^{m+1}})^{p^k}-1^{p^k})&=\nu_p(2^{2^{m+1}}-1)+k\nu_p(p)\\
&=\nu_p(2^{2^m}-1)+\nu_p(2^{2^m}+1)+k\\
&=k+1\\
\end{align}
\]
Therefore,
\[2^{2^{m+1}p^k}\equiv1\pmod{p^{k+1}}
\]
If \(\delta_{p^{k+1}}(2)<2^{m+1}p^k\),
\[\begin{align}
2^{2^{m+1}p^k}\equiv1\pmod{p^{k+1}}&\Longleftrightarrow\delta_{p^{k+1}}(2)\mid2^{m+1}p^k\\
&\Longleftrightarrow\delta_{p^{k+1}}(2)\mid2^mp^k\lor(k>0\land\delta_{p^{k+1}}(2)\mid2^{m+1}p^{k-1})\\
&\Longleftrightarrow2^{2^mp^k}\equiv1\pmod{p^{k+1}}\lor(k>0\land2^{2^{m+1}p^{k-1}}\equiv1\pmod{p^{k+1}})\\
\end{align}
\]
\[\begin{align}
p=2^{2^m}+1&\Longleftrightarrow2^{2^m}\equiv-1\pmod p\\
&\Longrightarrow2^{2^mp^k}\equiv-1\pmod p\\
&\Longleftrightarrow p\mid2^{2^mp^k}+1\\
&\Longrightarrow p\nmid2^{2^mp^k}-1\\
&\Longrightarrow p^{k+1}\nmid2^{2^mp^k}-1\\
&\Longleftrightarrow2^{2^mp^k}\not\equiv1\pmod{p^{k+1}}
\end{align}
\]
For \(k>0\),
\[\begin{align}
2^{2^{m+1}p^{k-1}}\equiv1\pmod{p^{k+1}}&\Longleftrightarrow p^{k+1}\mid2^{2^{m+1}p^{k-1}}-1\\
&\Longleftrightarrow\nu_p((2^{2^{m+1}})^{p^{k-1}}-1^{p^{k-1}})\ge k+1\\
&\Longleftrightarrow k\ge k+1\\
\end{align}
\]
Therefore,
\[2^{2^{m+1}p^{k-1}}\not\equiv1\pmod{p^{k+1}}
\]
Therefore,
\[2^{2^{m+1}p^k}\not\equiv1\pmod{p^{k+1}}
\]
By contradiction,
\[\delta_{p^{k+1}}(2)=2^{m+1}p^k
\]
Therefore, the smallest \(n\) such that \(2^n\equiv1\pmod{p^{k+1}}\) is \(2^{m+1}p^k\).
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