This sample GMAT Math question is a counting method problem solving question. The concept tested is to count the number of integers that are divisible by 3 can be formed based a criterion given in the question. An interesting GMAT 650 level permutation practice question.

Question 13: How many six-digit positive integers comprising only the digits 1 or 2 can be formed such that the number is divisible by 3?

- 3
- 20
- 22
- 38
- 360

@ INR

**Possibility 1: 111111.** All 6 digits are 1s. The sum of the digits is 6, which is divisible by 3.

Its digits can reorder in only one way. 1 number.

**Possibility 2: 222222.** All 6 digits are 2s. The sum of the digits is 12, which is divisible by 3.

The digits of 222222 can reorder in only one way. 1 number.

**Possibility 3: 111222.** __How did we arrive at this number?__

111111 is divisible by 3. If we remove three 1s, the sum will go down by 3. If we add three 2s, the sum will go up by 6. Because we subtracted a multiple of 3 and added a multiple of 3, the resultant sum is divisible by 3.

Ways in which 111222 can reorder = \\frac{6!}{3! × 3!}) = \\frac{720}{6 × 6}) = 20 Total number of positive integers that satisfy the conditions **= 1 + 1 + 20 = 22**

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