1956 AHSME Problems/Problem 34

Revision as of 22:07, 12 February 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem 34== If <math>n</math> is any whole number, <math>n^2(n^2 - 1)</math> is always divisible by <math> \textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 34

If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$

Solution

Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\boxed{\textbf{(A)} \quad 12}.$

-coolmath34

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png