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Difference between revisions of "2020 AIME I Problems/Problem 12"

m (→‎Solution 2 (Simpler, just basic mods and Fermat's theorem): fixed minor formatting error)
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Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
 
Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
  
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The divisibility criterion implies <math>149^n\equiv 2^n \pmod{5}</math>. This only occurs when <math>4 | n</math>, so let <math>n=4m</math>. We see <cmath>5 = v_5(149^{4m}-2^{4m}) = v_5((149^4)^{m}-(2^4)^{m}) = v_5(149^4-2^4)+v_5(m)</cmath> Since <math>149^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math>, then <math>v_5(149^4-2^4)=1</math>, so <math>v_5(m)=5</math>, hence <math>4 \cdot 5^4</math> divides <math>n</math>.  
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Now, multiplying <math>n</math> by <math>4</math>, we see <cmath>v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})</cmath> and since <math>149^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4n}-2^{4n})=1</math> meaning that we have that by LTE, <math>4 \cdot 5^4</math> divides <math>n</math>.  
  
 
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.
 
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.
  
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~kevinmathz + fireflame241
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~kevinmathz
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== Solution 2 (Simpler, just basic mods and Fermat's theorem)==
 
== Solution 2 (Simpler, just basic mods and Fermat's theorem)==
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BY THE WAY, please feel free to correct my formatting. I don't know latex.
  
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Note that for all n, <math>149^n - 2^n</math> is divisible by <math>149-2 = 147</math> because that is a factor. That is <math>3\cdot7^2</math>, so now we can clearly see that the smallest <math>n</math> to make the expression divisible by <math>3^3</math> is just <math>3^2</math>. Similarly, we can reason that the smallest n to make the expression divisible by <math>7^7</math> is just <math>7^5</math>.  
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Note that for all n, 149^n - 2^n is divisible by 149-2 = 147 because that is a factor. That is 3*7^2, so now we can clearly see that the smallest n to make the expression divisible by 3^3 is just 3^2. Similarly, we can reason that the smallest n to make the expression divisible by 7^7 is just 7^5.  
  
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Finally, for <math>5^5</math>, take mod <math>5</math> and mod <math>25</math> of each quantity (They happen to both be <math>-1</math> and <math>2</math> respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum <math>n</math> for divisibility by <math>5</math> is <math>4</math>, and other values are factors of <math>4</math>. Testing all of them (just <math>1,2,4</math> using mods-not too bad), <math>4</math> is indeed the smallest value to make the expression divisible by <math>5</math>, and this clearly is NOT divisible by <math>25</math>.
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Finally, for 5^5, take mod (5) and mod (25) of each quantity (They happen to both be -1 and 2 respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum n for divisibility by 5 is 4, and other values are factors of 4. Testing all of them(just 1,2,4 using mods-not too bad), 4 is indeed the smallest value to make the expression divisible by 5, and this clearly is NOT divisible by 25.
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Therefore, the smallest <math>n</math> to make this expression divisible by <math>5^5</math> is <math>2^2 \cdot 5^4</math>.
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Therefore, the smallest n to make this expression divisible by 5^5 is 2^2 * 5^4.
  
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Calculating the LCM of all these, one gets <math>2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Using the factor counting formula,
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Calculating the LCM of all these, one gets 2^2 * 3^2 * 5^4 * 7^5. Using the factor counting formula,
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the answer is <math>3\cdot 3\cdot 5\cdot 6</math> = <math>\boxed{270}</math>.
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the answer is 3*3*5*6 = <math>\boxed{270}</math>.
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Can someone please format better for me?
  
 
-Solution by thanosaops
 
-Solution by thanosaops

Revision as of 19:07, 16 March 2020

Problem

Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$

Solution 1

Lifting the Exponent shows that \[v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$. It also shows that \[v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$.


Now, multiplying $n$ by $4$, we see \[v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1$ meaning that we have that by LTE, $4 \cdot 5^4$ divides $n$.

Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.

~kevinmathz

Solution 2 (Simpler, just basic mods and Fermat's theorem)

BY THE WAY, please feel free to correct my formatting. I don't know latex.

Note that for all n, 149^n - 2^n is divisible by 149-2 = 147 because that is a factor. That is 3*7^2, so now we can clearly see that the smallest n to make the expression divisible by 3^3 is just 3^2. Similarly, we can reason that the smallest n to make the expression divisible by 7^7 is just 7^5.

Finally, for 5^5, take mod (5) and mod (25) of each quantity (They happen to both be -1 and 2 respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum n for divisibility by 5 is 4, and other values are factors of 4. Testing all of them(just 1,2,4 using mods-not too bad), 4 is indeed the smallest value to make the expression divisible by 5, and this clearly is NOT divisible by 25. Therefore, the smallest n to make this expression divisible by 5^5 is 2^2 * 5^4.

Calculating the LCM of all these, one gets 2^2 * 3^2 * 5^4 * 7^5. Using the factor counting formula, the answer is 3*3*5*6 = $\boxed{270}$. Can someone please format better for me?

-Solution by thanosaops

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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