Permutation and Divisiblity

Counting Methods Practice Question

An interesting GMAT problem solving practice question that combines two topics – counting methods and elementary number properties related to divisibility of numbers by 4.

Question

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

Correct Answer

Choice D. 60 such 5-digit numbers can be formed.

Explanatory Answer

What is to be computed?

We have to find the number of 5-digit numbers that can be formed using the digits 1, 2, 3 and 4.

Conditions to be met

  1. Each of the digits 1, 2, 3, and 4 should appear at least once in all the numbers formed.
  2. The numbers should be divisible by 4.

Next relevant question – what is the test of divisibility by 4?

A number is divisible by 4, if the last two digits (the rightmost two digits) of the number are divisible by 4.

Ever wondered why checking whether the last 2 digits are divisible by 4 will suffice?

What are the possible values for the last two digits if the number is divisible by 4 and is made of the digits 1,2, 3, and 4?

The last two digits can be 12, 24, 32, and 44.

Let us now compute the number of 5-digit numbers that can be formed for each of these scenarios.

Case 1: Ending with 12:
(a) The first three digits can be 2, 3, and 4. These 3 digits can rearrange in 3! = 6 ways
(b) The first three digits can be 1, 3, and 4. These 3 digits can rearrange in 3! = 6 ways
(c) The first three digits can be 3, 3, and 4. These 3 digits can rearrange in 3!/2! = 3 ways
(d) The first three digits can be 3, 4, and 4. These 3 digits can rearrange in 3!/2! = 3 ways
Total = 6 + 6 + 3 + 3 = 18 ways.

Note: We have to ensure that all 4 numbers 1, 2, 3, and 4 feature at least once in the numbers.

Case 2: Ending in 24:
Similar to case 1: 18 ways

Case 3: Ending in 32:
Similar to cases 1  and 2: 18 ways

Case 4: Ending in 44:
The first 3 digits have to be 1, 2, and 3 because we have to use each of the digits at least once in the numbers.
These 3 distinct digits can rearrange in 3! = 6 ways.

Total number of numbers that can be formed = 18 + 18 + 18 + 6 = 60.

The trap in this question is that we may extrapolate at the end of the case 1 that we will have 18 possibilities for each of the 4 cases and rush to mark 72 as the total number of numbers.

Queries, answers, comments welcome

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