Difference between revisions of "2021 AIME II Problems/Problem 15"
(→Problem) |
Dearpasserby (talk | contribs) (→Solution) |
||
Line 9: | Line 9: | ||
==Solution== | ==Solution== | ||
− | We | + | |
+ | Consider what happens when we try to calculate <math>f(n)</math> where n is not a square. If <math>k^2<n<(k+1)^2</math> for (positive) integer k, recursively calculating the value of the function gives us <math>f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n</math>. Note that this formula also returns the correct value when <math>n=(k+1)^2</math>, but not when <math>n=k^2</math>. Thus <math>f(n)=k^2+3k+2-n</math> for <math>k^2<n \leq (k+1)^2</math>. | ||
+ | |||
+ | If <math>2 \mid (k+1)^2-n</math>, <math>g(n)</math> returns the same value as <math>f(n)</math>. This is because the recursion once again stops at <math>(k+1)^2</math>. We seek a case in which <math>f(n)<g(n)</math>, so obviously this is not what we want. We want <math>(k+1)^2,n</math> to have a different parity, or <math>n, k</math> have the same parity. When this is the case, <math>g(n)</math> instead returns <math>(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n</math>. | ||
+ | |||
+ | Write <math>7f(n)=4g(n)</math>, which simplifies to <math>3k^2+k-10=3n</math>. Notice that we want the <math>LHS</math> expression to be divisible by 3; as a result, <math>k \equiv 1 \pmod{3}</math>. We also want n to be strictly greater than <math>k^2</math>, so <math>k-10>0, k>10</math>. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is <math>k=16</math>, giving <math>n=258</math>. | ||
+ | |||
+ | Indeed, <math>f(258)=48, g(258)=84</math>, so we're done. | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2021|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:19, 22 March 2021
Problem
Let and be functions satisfying and for positive integers . Find the least positive integer such that .
Solution
Consider what happens when we try to calculate where n is not a square. If for (positive) integer k, recursively calculating the value of the function gives us . Note that this formula also returns the correct value when , but not when . Thus for .
If , returns the same value as . This is because the recursion once again stops at . We seek a case in which , so obviously this is not what we want. We want to have a different parity, or have the same parity. When this is the case, instead returns .
Write , which simplifies to . Notice that we want the expression to be divisible by 3; as a result, . We also want n to be strictly greater than , so . The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is , giving .
Indeed, , so we're done.
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.