1951 AHSME Problems/Problem 37

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 $2520-1=\boxed{\textbf{(D)}\ 2519}$

See Also

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