Difference between revisions of "2017 AIME II Problems/Problem 8"
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− | + | ==Problem== | |
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. | ||
− | <math>\ | + | ==Solution 1== |
− | <math>\boxed{134}</math> | + | We start with the last two terms of the polynomial <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</math>, which are <math>\frac{n^5}{5!}+\frac{n^6}{6!}</math>. This can simplify to <math>\frac{6n^5+n^6}{720}</math>, which can further simplify to <math>\frac{n^5(6+n)}{720}</math>. Notice that the prime factorization of <math>720</math> is <math>5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2</math>. In order for <math>\frac{n^5(6+n)}{720}</math> to be an integer, one of the parts must divide <math>5, 3</math>, and <math>2</math>. Thus, one of the parts must be a multiple of <math>5, 3</math>, and <math>2</math>, and the LCM of these three numbers is <math>30</math>. This means <cmath>n^5 \equiv 0\pmod{30}</cmath> or <cmath>6+n \equiv 0\pmod{30}</cmath> Thus, we can see that <math>n</math> must equal <math>0\pmod{30}</math> or <math>-6\pmod{30}</math>. Note that as long as we satisfy <math>\frac{6n^5+n^6}{720}</math>, <math>2!, 3!</math>, and <math>4!</math> will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. <math>4! = 2\cdot2\cdot2\cdot2\cdot3</math>, and this will be divisible by <math>2^4\cdot3^4\cdot5^4</math>. Now, since we know that <math>n</math> must equal <math>0\pmod{30}</math> or <math>-6\pmod{30}</math> in order for the polynomial to be an integer, <math>n\equiv0, 24\pmod{30}</math>. To find how many integers fulfill the equation and are <math><2017</math>, we take <math>\left \lfloor\frac{2017}{30} \right \rfloor</math> and multiply it by <math>2</math>. Thus, we get <math>67\cdot2=\boxed{134}</math>. |
+ | |||
+ | ~Solution by IronicNinja~ | ||
+ | |||
+ | ==Solution 2== | ||
+ | Taking out the <math>1+n</math> part of the expression and writing the remaining terms under a common denominator, we get <math>\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)</math>. Therefore the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> must equal <math>720m</math> for some positive integer <math>m</math>. | ||
+ | Taking both sides mod <math>2</math>, the result is <math>n^6 \equiv 0 \pmod{2}</math>. Therefore <math>n</math> must be even. If <math>n</math> is even, that means <math>n</math> can be written in the form <math>2a</math> where <math>a</math> is a positive integer. Replacing <math>n</math> with <math>2a</math> in the expression, <math>64a^6+192a^5+480a^4+960a^3+1440a^2</math> is divisible by <math>16</math> because each coefficient is divisible by <math>16</math>. Therefore, if <math>n</math> is even, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> is divisible by <math>16</math>. | ||
+ | |||
+ | Taking the equation <math>n^6+6n^5+30n^4+120n^3+360n^2=720m</math> mod <math>3</math>, the result is <math>n^6 \equiv 0 \pmod{3}</math>. Therefore <math>n</math> must be a multiple of <math>3</math>. If <math>n</math> is a multiple of three, that means <math>n</math> can be written in the form <math>3b</math> where <math>b</math> is a positive integer. Replacing <math>n</math> with <math>3b</math> in the expression, <math>729b^6+1458b^5+2430b^4+3240b^3+3240b^2</math> is divisible by <math>9</math> because each coefficient is divisible by <math>9</math>. Therefore, if <math>n</math> is a multiple of <math>3</math>, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> is divisibly by <math>9</math>. | ||
+ | |||
+ | Taking the equation <math>n^6+6n^5+30n^4+120n^3+360n^2=720m</math> mod <math>5</math>, the result is <math>n^6+n^5 \equiv 0 \pmod{5}</math>. The only values of <math>n (\text{mod }5)</math> that satisfy the equation are <math>n\equiv0(\text{mod }5)</math> and <math>n\equiv4(\text{mod }5)</math>. Therefore if <math>n</math> is <math>0</math> or <math>4</math> mod <math>5</math>, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> will be a multiple of <math>5</math>. | ||
+ | |||
+ | The only way to get the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> to be divisible by <math>720=16 \cdot 9 \cdot 5</math> is to have <math>n \equiv 0 \pmod{2}</math>, <math>n \equiv 0 \pmod{3}</math>, and <math>n \equiv 0 \text{ or } 4 \pmod{5}</math>. By the Chinese Remainder Theorem or simple guessing and checking, we see <math>n\equiv0,24 \pmod{30}</math>. Because no numbers between <math>2011</math> and <math>2017</math> are equivalent to <math>0</math> or <math>24</math> mod <math>30</math>, the answer is <math>\frac{2010}{30}\times2=\boxed{134}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | -TheUltimate123 | ||
+ | |||
+ | ==Step Solution== | ||
+ | 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 \pmod5</math>, so combining with divisibility by 6, <math>n</math> is <math>24</math> or <math>0\pmod{30}</math>. There are <math>67</math> cases for each, hence the answer <math>\boxed{134}</math>. | ||
+ | |||
+ | =See Also= | ||
+ | {{AIME box|year=2017|n=II|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:29, 18 October 2022
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.