How many times will the digit 5 come in counting from 1 to 99, excluding those numbers which are divisible by 3 ?
16
15
14
13
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Next, identify the numbers between 1 and 99 that contain the digit 5 and are divisible by 3. These numbers are:
– 15 (1+5=6, divisible by 3) – contains one 5
– 45 (4+5=9, divisible by 3) – contains one 5
– 51 (5+1=6, divisible by 3) – contains one 5
– 54 (5+4=9, divisible by 3) – contains one 5
– 57 (5+7=12, divisible by 3) – contains one 5
– 75 (7+5=12, divisible by 3) – contains one 5
These are 6 numbers. Let’s count the occurrences of the digit 5 *within* these numbers: 15 (one 5), 45 (one 5), 51 (one 5), 54 (one 5), 57 (one 5), 75 (one 5). Total occurrences of 5 in these numbers are 6.
The numbers containing 5 are: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85, 95.
The numbers from this list that are divisible by 3 are: 15, 45, 51, 54, 57, 75.
We need to count the 5s in the numbers *not* divisible by 3: 5, 25, 35, 50, 52, 53, 55, 56, 58, 59, 65, 85, 95.
Counting the 5s in this remaining list:
5 (one 5), 25 (one 5), 35 (one 5), 50 (one 5), 52 (one 5), 53 (one 5), 55 (two 5s), 56 (one 5), 58 (one 5), 59 (one 5), 65 (one 5), 85 (one 5), 95 (one 5).
Total count = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 16.