# GMAT Challenging Question 14 | Counting Methods

#### GMAT Sample Questions | Permutation Combination

The given question is GMAT 700 level problem solving question combining concepts in divisibility of numbers (number properties) and counting methods (permutation combination). The question highlights the concept of listing down all the different possibilities that satisfy the condition mentioned about the 4-digit number and then counting them to arrive at the answer.

Question 14: How many four-digit positive integers exist that contain the block 25 and are divisible by 75. (2250 and 2025 are two such numbers)?

1. 90
2. 63
3. 34
4. 87
5. 62

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### Explanatory Answer | GMAT Counting Methods

#### Question 1: What will be the form of such 4-digit numbers?

The 4-digit numbers should contain the block 25.
The required 4-digit numbers will be of the form:
a. 25_ _
b. _ 25 _
c. _ _ 25

#### Question 2: What is the test of divisibility by 75?

If a number is divisible by 75, then it will be divisible by 25 and 3.

#### Count the number of 4-digit numbers for three possiblities

a. Numbers of the form 25 _ _ that are divisible by 75
A number of the form 25_ _ is divisible by 25 if its rightmost 2 digits are 00, 25, 50, or 75.

Check which of these numbers are also multiples of 3
Only one number, 2550 satisfies the condition.

b. Numbers of the form _ 25 _ that are divisible by 75
A number of the form _ 25 _ is divisible by 25 if its unit digit is 0.
The 4-digit number will be of the form _ 250

What options exist for the left most digit so that the number is also divisible by 3?
The sum of the right most 3 digits of the number = 2 + 5 + 0 = 7.
If the first digit is 2 or 5 or 8, the sum of the 4 digits will be divisible by 3.
There are three 4-digit numbers that match the form _ 25 _ and are divisible by 75.

c. Numbers of the form _ _ 25 that are divisible by 75
All numbesr of the form _ _ 25 is divisible by 25.

What options exist for the first 2 digits so that the number is also divisible by 3?
We already have a 2 and 5 whose sum is 7. 7 is a multiple of 3 plus 1.
We have a (3k + 1) with us. If we add a (3m + 2), the sum will be 3(k + m) + 1 + 2 = 3(k + m) + 3, which is divisible by 3.
The least 2 digit number that is of the form (3m + 2) is 11.
For example, if 11 takes the 1st 2 places, the number is divisible by 3

11 is not the only such number.
All 2-digit numbers of the form (3m + 2) will work

How many are there? The largest 2-digit number that is of the form 3m + 2 is 98.
And all of these numbers are in arithmetic progression with a common difference of 3
So, apply the arithmetic progression formula to compute the nth term: 98 = 11 + (n - 1)3
3(n - 1) = 87
(n - 1) = 29
Or n = 30
30 such 4-digit numbers exist

#### Add the count of all three possiblities

1 + 3 + 30 = 34
34 such 4-digit numbers exist

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