Difference between revisions of "2017 AIME II Problems/Problem 8"
(→Solution 1) |
m (→Solution 1) |
||
Line 2: | Line 2: | ||
Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer. | Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer. | ||
− | ==Solution 1== | + | ==Solution 1 (Not Rigorous)== |
Writing the last two terms with a common denominator, we have <math>\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}</math> By inspection. this yields that <math>n \equiv 0, 24 \pmod{30}</math>. Therefore, we get the final answer of <math>67 + 67 = \boxed{134}</math>. | Writing the last two terms with a common denominator, we have <math>\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}</math> By inspection. this yields that <math>n \equiv 0, 24 \pmod{30}</math>. Therefore, we get the final answer of <math>67 + 67 = \boxed{134}</math>. | ||
Revision as of 21:41, 21 February 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 .
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.