Problem Solving

Max@Math Revolution wrote:[GMAT math practice question]

When n is divided by 2, 3, 5, 7 and 9, the remainder is 1. What is the smallest value of the positive integer n?

A. 100
B. 211
C. 421
D. 631
E. 841

Since the division of n by 2, 3, 5, 7 and 9 produces a remainder of 1, n - 1 is divisible by 2, 3, 5, 7 and 9. The LCM of these numbers will give us the smallest value of n - 1.

The LCM of 2, 3, 5, 7, and 9 is:

2 x 3^2 x 5 x 7 = 630, so we see that 631 is the smallest value that will have a reminder of 1 when divided by 2, 3, 5, 7 and 9.

Answer: D

To add a bit to this smart approach by GMATGuruNY, why choose 9 as the divisor to check? Because it's very easy to tell when a number is a multiple of 9. If the sum of the digits of the number is 9 or a multiple of 9, then the number itself is a multiple of 9.

So after eliminating A because it's even and therefore a multiple of 2 (no remainder), eliminating B, C, and E becomes quite simple.

B) 211-1 = 210. Since 2+1+0 = 3, and 3 is not a multiple of 9, you know that 210 is not a multiple of 9. Eliminate.
C) 421-1 = 420. Since 4+2+0 = 6, and 6 is not a multiple of 9, you know that 420 is not a multiple of 9. Eliminate.
D) 631 -1 = 630. Since 6+3+0=9, and 9 is a multiple of 9, you know that 630 IS a multiple of 9. Keep.
E) 841-1=840. Since 8+4+0=12, and 12 is not a multiple of 9, you know that 840 is not a multiple of 9. Eliminate.

GMATGuruNY wrote:

Max@Math Revolution wrote:[GMAT math practice question]

When n is divided by 2, 3, 5, 7 and 9, the remainder is 1. What is the smallest value of the positive integer n?

A. 100
B. 211
C. 421
D. 631
E. 841

Since n must yield a remainder when divided by 2, n cannot be even.
Eliminate A.

To yield a remainder of 1 when divided by 9, n must be equal to ONE MORE THAN A MULTIPLE OF 9.
Thus, the correct answer choice must be equal to (MULTIPLE OF 9) + 1:
B) 211 = 210 + 1 --> Nope.
C) 421 = 420 + 1 --> Nope.
D) 631 = 630 + 1 --> Success!
E) 841 = 840 + 1 --> Nope.

The correct answer is D.