2025 AMC 12A Problem 7

Attempt Problem 7 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

In a certain alien world, the maximum running speed vv of an organism is dependent on its number of toes nn and number of eyes m.m. The relationship can be expressed as v=knambv = k n^a m^b centimeters per hour, where k,k, a,a, and bb are integer constants. In a population where all organisms have 55 toes, log⁡v=4+2log⁡m;\log v = 4 + 2\log m; and in a population where all organisms have 2525 eyes, log⁡v=4+4log⁡n,\log v = 4 + 4\log n, where the logarithms are base 10.10. What is k+a+b?k + a + b?
2020
2121
2222
2323
2424
Answer: C
Concepts:logarithmsystem of equations
Small hint:
Take logs: log⁡v=log⁡k+alog⁡n+blog⁡m\log v = \log k + a\log n + b\log m
Big hint:
Setting n=5n = 5 gives b=2b = 2 and log⁡k+alog⁡5=4\log k + a\log 5 = 4; setting m=25m = 25 gives a=4a = 4
Solution:
Taking logarithms, log⁡v=log⁡k+alog⁡n+blog⁡m.\log v = \log k + a\log n + b\log m.
With n=5,n = 5, this reads log⁡v=(log⁡k+alog⁡5)\log v = (\log k + a\log 5) +blog⁡m,+ b\log m, matching 4+2log⁡m,4 + 2\log m, so b=2b = 2 and log⁡k+alog⁡5=4.\log k + a\log 5 = 4.
With m=25,m = 25, it reads log⁡v=(log⁡k+blog⁡25)\log v = (\log k + b\log 25) +alog⁡n,+ a\log n, matching 4+4log⁡n,4 + 4\log n, so a=4a = 4 and log⁡k+2log⁡25=4.\log k + 2\log 25 = 4.
Then log⁡k=4−log⁡625\log k = 4 - \log 625 =log⁡10000625= \log\dfrac{10000}{625} =log⁡16,= \log 16, so k=16.k = 16. Hence k+a+b=16+4+2=22.k + a + b = 16 + 4 + 2 = 22.
Thus, the correct answer is C.
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