Difference between revisions of "2017 AIME II Problems/Problem 8"
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− | Clearly <math>1+n</math> is an integer. The part we need to verify as an integer is, upon common denominator, <math>\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}</math>. Clearly, the numerator must be even for the fraction to be an integer. Therefore, <math>n^6</math> is even and n is even, aka <math>n=2k</math> for some integer <math>k</math>. Then, we can substitute <math>n=2k</math> and see that <math>\frac{n^2}{2}</math> is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get <math>\frac{60k^3+30k^4+12k^5+4k^6}{45}</math>. It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the <math>4k^6</math>, and we see that <math>k=3b</math> for some integer <math>b</math>. From there we now know that <math>n=6b</math>. If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that <math>n^5(6n+1) \ | + | Clearly <math>1+n</math> is an integer. The part we need to verify as an integer is, upon common denominator, <math>\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}</math>. Clearly, the numerator must be even for the fraction to be an integer. Therefore, <math>n^6</math> is even and n is even, aka <math>n=2k</math> for some integer <math>k</math>. Then, we can substitute <math>n=2k</math> and see that <math>\frac{n^2}{2}</math> is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get <math>\frac{60k^3+30k^4+12k^5+4k^6}{45}</math>. It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the <math>4k^6</math>, and we see that <math>k=3b</math> for some integer <math>b</math>. From there we now know that <math>n=6b</math>. If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that <math>n^5(6n+1) \equiv 0(\mod 5)</math>, so combining with divisibility by 6, <math>n</math> is <math>24</math> or <math>0</math> (mod <math>30</math>). There are <math>67</math> cases for each, hence the answer <math>\boxed{134}</math>. |
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=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 18:06, 3 March 2018
Contents
[hide]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
Step Solution
Clearly is an integer. The part we need to verify as an integer is, upon common denominator, . Clearly, the numerator must be even for the fraction to be an integer. Therefore, is even and n is even, aka for some integer . Then, we can substitute and see that is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get . It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the , and we see that for some integer . From there we now know that . If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that , so combining with divisibility by 6, is or (mod ). There are cases for each, hence the answer .
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.