1951 AHSME Problems/Problem 37

Revision as of 16:27, 19 September 2022 by Lucolas (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 37

A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves a remainder of $8$, by $8$ leaves a remainder of $7$, etc., down to where, when divided by $2$, it leaves a remainder of $1$, is:

$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$

Solution

If we add $1$ to the number, it becomes divisible by $10, 9, 8, \cdots, 2, 1$. The LCM of $1$ through $10$ is $2520$, therefore the number we want to find is $2520-1=\boxed{\textbf{(D)}\ 2519}$

Video Solution

https://youtu.be/UQGamWE1XOo

~Lucas

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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