Counting Methods : GMAT Problem Solving

1. The number is a 4 – digit number.
2. It is an odd number.
3. It is a multiple of 5.
4. 3 cannot be a digit in the number.

Step 1: Let us start with the units place – the right most digit of the number

Any number that is divisible by 5 will end in 0 or 5. Combining this information with condition 2 that the number is an odd number, we can conclude that the unit digit of all these numbers is 5.

Therefore, the unit digit has only one option. That was easy. One out of the 4 digits out of our way.

Step 2: Let us find out the number of options for the left most digit

The thousands place, the left most digit of these numbers cannot be 0.
The question stipulates that 3 cannot be a digit of the number. So, it cannot be 3.
Out of the total 10 possibilities, 0 to 9, we cannot use 2 digits.
That leaves us with 8 options for the thousands or the left most digit.

Step 3: The remaining two digits – hundreds and tens place

a. The hundreds place cannot be 3. It can take any value from 0 to 9 except 3. So, it has 9 options.
b. The tens place also cannot be 3. Using the same argument that we used for hundreds place, the tens place of these numbers has 9 options.

Therefore, the number of 4 digit numbers that can be formed that satisfy all of the mentioned conditions

= 8 * 9 * 9 * 1 = 648 numbers.