Practice
Find \(v_3(10^{2025}-1)\).
Use \(3\mid10-1\).
By LTE, \(v_3(10^{2025}-1)=v_3(10-1)+v_3(2025)=v_3(9)+4=2+4=6\).
Find \(v_7(8^{100}-1)\).
\(7\mid8-1\).
By LTE, \(v_7(8^{100}-1)=v_7(7)+v_7(100)=1+0=1\).
Find \(v_5(6^{50}-1)\).
\(5\mid6-1\).
\(v_5(6^{50}-1)=v_5(5)+v_5(50)=1+2=3\).
Find \(v_3(2^{100}-1)\).
\(3\mid2+1\), and \(100\) is even.
Since \(3\mid2+1\) and \(100\) is even, \(v_3(2^{100}-1)=v_3(3)+v_3(100)=1\).
Find \(v_2(3^{100}-1)\).
Use the formula for \(p=2\).
\(v_2(3^{100}-1)=v_2(2)+v_2(4)+v_2(100)-1=1+2+2-1=4\).
Find all positive integers \(n\) such that \(7^3\mid8^n-1\).
Compute \(v_7(8^n-1)\).
\(v_7(8^n-1)=1+v_7(n)\). We need \(1+v_7(n)\ge3\), i.e. \(v_7(n)\ge2\). Answer: all \(n\) divisible by \(49\).
Find the largest \(k\) such that \(9^k\mid10^{2025}-1\).
First find \(v_3\).
From Problem 1, \(v_3(10^{2025}-1)=6\). Since \(9^k=3^{2k}\), \(2k\le6\), hence \(k=3\).
Find \(v_5(11^n-1)\).
\(5\mid11-1\).
By LTE, \(v_5(11^n-1)=v_5(10)+v_5(n)=1+v_5(n)\).
Find \(v_3(4^n-1)\).
\(3\mid4-1\).
\(v_3(4^n-1)=v_3(3)+v_3(n)=1+v_3(n)\).
Find \(v_3(5^{2n}-1)\).
\(3\mid5+1\), and the exponent \(2n\) is even.
By the \(a+b\) case, \(v_3(5^{2n}-1)=v_3(6)+v_3(2n)=1+v_3(n)\).
Find \(v_2(5^n-1)\) depending on the parity of \(n\).
For odd \(n\), use the \(5-1\) factor; for even \(n\), use the formula for \(2\).
If \(n\) is odd, \(v_2(5^n-1)=v_2(5-1)=2\). If \(n\) is even, \(v_2(5^n-1)=v_2(4)+v_2(6)+v_2(n)-1=2+1+v_2(n)-1=2+v_2(n)\).
Find all \(n\) such that \(16\mid3^n-1\).
Split by parity of \(n\).
If \(n\) is odd, \(v_2(3^n-1)=v_2(2)=1\), not enough. If \(n\) is even, then \(v_2(3^n-1)=v_2(2)+v_2(4)+v_2(n)-1=2+v_2(n)\). We need \(2+v_2(n)\ge4\), so \(v_2(n)\ge2\). Answer: \(4\mid n\).
Find all \(n\) such that \(3^5\mid4^n-1\).
Use \(v_3(4^n-1)=1+v_3(n)\).
We need \(1+v_3(n)\ge5\), i.e. \(v_3(n)\ge4\). Answer: \(81\mid n\).
Let \(p\) be an odd prime, \(p\mid a-b\), \(p\nmid ab\). Prove that \(p^r\mid a^{p^{r-1}}-b^{p^{r-1}}\) for every \(r\ge1\).
Apply LTE to the exponent \(p^{r-1}\).
By LTE, \(v_p(a^{p^{r-1}}-b^{p^{r-1}})=v_p(a-b)+v_p(p^{r-1})\). Since \(p\mid a-b\), the first term is at least \(1\), and the second equals \(r-1\). The total exponent is at least \(r\), so divisibility by \(p^r\) follows.
Find the largest \(k\) such that \(3^k\mid2^{2n}-1\).
Replace \(2^{2n}\) by \(4^n\).
\(2^{2n}-1=4^n-1\). Since \(3\mid4-1\), \(v_3(4^n-1)=1+v_3(n)\). Therefore \(k=1+v_3(n)\).
Find \(v_3(13^n-1)\).
\(3\mid13-1\).
By LTE, \(v_3(13^n-1)=v_3(12)+v_3(n)=1+v_3(n)\).
Find all positive integers \(n\) such that \(3^n\mid10^n-1\).
Compare \(n\) with \(v_3(10^n-1)\).
By LTE, \(v_3(10^n-1)=v_3(9)+v_3(n)=2+v_3(n)\). We need \(2+v_3(n)\ge n\). The values \(n=1,2,3\) work. If \(n\ge4\), then \(2+v_3(n)
Find all positive integers \(n\) such that \(7^n\mid8^n-1\).
Use \(v_7(8^n-1)=1+v_7(n)\).
We need \(1+v_7(n)\ge n\). For \(n=1\), the condition holds. If \(n\ge2\), then \(v_7(n)\le \log_7 n
Let \(p\) be an odd prime, \(p\mid a-b\), \(p^2\nmid a-b\), and \(p\nmid ab\). Prove that \(p^r\mid a^n-b^n\) if and only if \(p^{r-1}\mid n\).
Under the conditions, \(v_p(a-b)=1\).
By LTE, \(v_p(a^n-b^n)=v_p(a-b)+v_p(n)=1+v_p(n)\). The divisibility \(p^r\mid a^n-b^n\) is equivalent to \(1+v_p(n)\ge r\), i.e. \(v_p(n)\ge r-1\). This is equivalent to \(p^{r-1}\mid n\).
Prove that for \(m\ge1\), \(v_2(3^{2^m}-1)=m+2\).
Apply the \(p=2\) LTE formula with \(a=3\), \(b=1\), \(n=2^m\).
\(v_2(3^{2^m}-1)=v_2(3-1)+v_2(3+1)+v_2(2^m)-1=1+2+m-1=m+2\).