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?

- 120
- 24
- 72
- 60
- 54

From INR

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.

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**

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