Difference between revisions of "2017 AIME II Problems/Problem 8"
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Note that <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}</math> will have a denominator that divides <math>5!</math>. Therefore, for the expression to be an integer, <math>\frac{n^6}{6!}</math> must have a denominator that divides <math>5!</math>. Thus, <math>6\mid n^6</math>, and <math>6\mid n</math>. Let <math>n=6m</math>. Substituting gives <math>1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math>. Note that the first <math>5</math> terms are integers, so it suffices for <math>\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math> to be an integer. This simplifies to <math>\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)</math>. It follows that <math>5\mid m^5(m+1)</math>. Therefore, <math>m</math> is either <math>0</math> or <math>4</math> modulo <math>5</math>. However, we seek the number of <math>n</math>, and <math>n=6m</math>. By CRT, <math>n</math> is either <math>0</math> or <math>24</math> modulo <math>30</math>, and the answer is <math>67+67=\boxed{134}</math>. | Note that <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}</math> will have a denominator that divides <math>5!</math>. Therefore, for the expression to be an integer, <math>\frac{n^6}{6!}</math> must have a denominator that divides <math>5!</math>. Thus, <math>6\mid n^6</math>, and <math>6\mid n</math>. Let <math>n=6m</math>. Substituting gives <math>1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math>. Note that the first <math>5</math> terms are integers, so it suffices for <math>\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math> to be an integer. This simplifies to <math>\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)</math>. It follows that <math>5\mid m^5(m+1)</math>. Therefore, <math>m</math> is either <math>0</math> or <math>4</math> modulo <math>5</math>. However, we seek the number of <math>n</math>, and <math>n=6m</math>. By CRT, <math>n</math> is either <math>0</math> or <math>24</math> modulo <math>30</math>, and the answer is <math>67+67=\boxed{134}</math>. | ||
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+ | -TheUltimate123 | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=7|num-a=9}} | {{AIME box|year=2017|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:49, 3 March 2018
Problem
Find the number of positive integers less than such that is an integer.
Solution 1 (Not Rigorous)
Writing the last two terms with a common denominator, we have By inspection. this yields that . Therefore, we get the final answer of .
Solution 2
Taking out the part of the expression and writing the remaining terms under a common denominator, we get . Therefore the expression must equal for some positive integer . Taking both sides mod , the result is . Therefore must be even. If is even, that means can be written in the form where is a positive integer. Replacing with in the expression, is divisible by because each coefficient is divisible by . Therefore, if is even, is divisible by .
Taking the equation mod , the result is . Therefore must be a multiple of . If is a multiple of three, that means can be written in the form where is a positive integer. Replacing with in the expression, is divisible by because each coefficient is divisible by . Therefore, if is a multiple of , is divisibly by .
Taking the equation mod , the result is . The only values of that satisfy the equation are and . Therefore if is or mod , will be a multiple of .
The only way to get the expression to be divisible by is to have , , and . By the Chinese Remainder Theorem or simple guessing and checking, we see . Because no numbers between and are equivalent to or mod , the answer is .
Solution 3
Note that will have a denominator that divides . Therefore, for the expression to be an integer, must have a denominator that divides . Thus, , and . Let . Substituting gives . Note that the first terms are integers, so it suffices for to be an integer. This simplifies to . It follows that . Therefore, is either or modulo . However, we seek the number of , and . By CRT, is either or modulo , and the answer is .
-TheUltimate123
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.