Chapter 15.5, Problem 16WE

To determine

To calculate: The number of multiples of 3 less than 1000 with digits 2, 5 and 9.

Answer to Problem 16WE

The number of multiples of 3 less than 1000 with digits 2, 5 and 9 are $11$ .

Explanation of Solution

Given information:

The number of multiples of 3 less than 1000 with digits 2, 5 and 9 are to be formed.

Formula used:

Fundamental counting principle states that if there are m ways to make a selection and n ways to make a second selection then there are $m\times n$ number of ways in which two selection can be made.

Calculation:

Consider the provided information that number of multiples of 3 less than 1000 with digits 2, 5 and 9 are to be formed.

A number is a multiple of 3 if the sum of digits is a multiple of 3.

From the provided digits no single digit multiple of 3 is there.

Only 1 digit number can be formed out of 3 provided digits that is 9 which is a multiple of 3.

Let 2-digit number be $\boxed{}\boxed{}$ .

If 2 is fixed at ten’s digit then there are no possible numbers at one’s digit as sum of digits won’t be a multiple of 3.

Similarly with 5 at ten’s digit.

When 9 is fixed at ten’s digit then possible numbers at one’s digit is 9 since sum of digits of 99 add up to multiple of 3 that is 18.

So, numbers that are multiple of 9 and 99 till now.

Let 3-digit number be $\boxed{}\boxed{}\boxed{}$ .

If 2 is fixed at hundredth’s place then numbers at ten’s and one’s digit are 2 and 5 since sum of digits of 225 add up to multiple of 3 that is 9.

Now permutations of number 225 is also amultiple of 3.

That is 252 and 522.

If 5 is fixed at hundredth’s place then numbers at ten’s and one’s digit are 2 and 5 since sum of digits of 525 add up to multiple of 3 that is 12.

Now permutations of number 525 is also a multiple of 3.

That is 552 and 255.

Now the remaining multiples of 3 are 222, 555 and 999.

Count the number of multiples of 3 obtained above.

The count is 11.

Thus, multiples of 3 less than 1000 with digits 2, 5 and 9 are $11$ .