GMAT Sample Questions | Permutation Q14

Numbers and Digits | Permutation Combination

This GMAT quant question is a combinatorics problem solving question. The concept tested is to find the number of ways the digits of a number can be rearranged after factoring a divisibility constraint. A medium difficulty permutation question.

Question 14: How many five-digit positive integers comprising only the digits 1, 2, 3, and 4, each appearing at least once, exist such that the number is divisible by 4?

  1. 120
  2. 24
  3. 72
  4. 60
  5. 54

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Explanatory Answer | GMAT Numbers and Digits

Step 1 of solving this GMAT Permutation Question: List Down the Constraints

The conditions that the number should satisfy
1. It is a 5-digit number
2. Digits should comprise only 1, 2, 3, and 4.
3. Each digit should appear at least once in the 5-digit number.
4. The number is divisible by 4

Test of divisibility for 4
The last two digits (rightmost two digits) should be divisible by 4.

Step 2 of solving this GMAT Permutation Question: Applying constraints, one at a time

Let's start with constraints 1, 2, and 4.
What all possible values can the rightmost two digits be such that the number is divisible by 4 and should comprise only 1, 2, 3, and 4?
Possible values for the last two digits: 12, 24, 32, and 44

Number of numbers ending with 12 (all 4 digits should appear at least once)
Case 1: The first 3 digits are 3, 4, and 1. i.e., 3 4 112. The first 3 digits can reorder in 3! ways = 6.
Case 2: The first 3 digits are 3, 4, and 2. i.e., 3 4 212. The first 3 digits can reorder in 3! ways = 6.
Case 3: The first 3 digits are 3, 4, and 3. i.e., 3 4 312. The first 3 digits can reorder in \\frac{3!}{2!}) ways = 3.
Case 4: The first 3 digits are 3, 4, and 4. i.e., 3 4 412. The first 3 digits can reorder in \\frac{3!}{2!}) ways = 3.
Number of numbers ending with 12 = 18.

Similarly, the numbers ending with 24 and 32 occur 18 times each.

Number of numbers ending in 44 The first 3 digits have to be 1, 2, and 3.
These 3 digits can reorder in 3! = 6 ways.
So, 6 numbers end in 44.

Total required 5-digit numbers 18 + 18 + 18 + 6 = 60

Choice D is the correct answer.



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