1951 AMC 12 Problem 37

Question

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:

59

419

1259

2519

None of these answers

Adding

1

to the desired number removes every listed remainder

Find

\operatorname{lcm}(2,3,\ldots,10)

If the number is

N,

then

N+1

is divisible by every integer from

2

through

10.

Their least common multiple is

2^3\cdot3^2\cdot5\cdot7=2520.

Thus the least positive number fitting all the conditions is

N=2520-1=2519.

Thus, the correct answer is D.

Problem 37 in other years