2017 AIME II Problems/Problem 8
Problem
Find the number of positive integers less than such that is an integer.
Solution 1
We start with the last two terms of the polynomial , which are . This can simplify to , which can further simplify to . Notice that the prime factorization of is . In order for to be an integer, one of the parts must divide , and . Thus, one of the parts must be a multiple of , and , and the LCM of these three numbers is . This means or Thus, we can see that must equal or . Note that as long as we satisfy , , and will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. , and this will be divisible by . Now, since we know that must equal or in order for the polynomial to be an integer, . To find how many integers fulfill the equation and are , we take and multiply it by . Thus, we get .
Solution by IronicNinja~
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 . 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 |
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