Eliminating Answers Methodsukhman wrote:What is the smallest number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder 8, when divided by 8 leaves a remainder 7, and so on till dividing by 2 leaves a remainder 1.
(A) 2529
(B) 1529
(C) 3509
(D) 3259
(E) 2519
3500 is a multiple of 7. So dividing 3509 by 7 leaves a remainder of 2. Eliminate C.
3200 is a multiple of 8, and 59 is 3 more than a multiple of 8. So dividing 3259 by 8 leaves a remainder of 3. Eliminate D.
The digits of 2529 add up to 18, which is divisible by 3. So 2529 is divisible by 3. Eliminate A.
1500 is divisible by 4, and 29 is one more than a multiple of 4. So eliminate B.
The correct answer must be E.
Elegant Method
Find the least common multiple of 10, 9, 8, 7, 6, 5, 4, 3 and 2.
10 x 9 x 7 x 4 captures all of the primes.
10 x 9 x 7 x 4 = 90 x 28 = (100 x 28) - 280 = 2520.
Subtract 1 to get a number 1 less than a multiple of any of those numbers.
2520 - 1 = 2519
The correct answer is E.
