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Asked by  | 18th Sep, 2012, 11:15: AM

Expert Answer:

The hundreds place of a three digit number can never be zero.

So, the hundreds place of the required three-digit number can be filled by any of the four digits 1, 2, 3 and 4.

That is hundreds place digit can be filled by 4 ways.

 

A number is divisible by 3, if sum of its digits is divisible by 3.

 

If the hundreds place digit is either 1 or 4, then the sum of tens and ones place digits must be of the form 3x + 2, where x is a whole number. After fixing 1 or 4 at hundreds place, the tens and ones place of the number are any of the following:

02, 11, 14, 20, 23, 32, 41, 44

In such case 2 × 8 = 16 numbers of such 3-digit numbers are formed.

 

If the hundreds place digit is 2, then the sum of tens and ones place digits must be of the form 3y + 1, where y is a whole number. After fixing 2 at hundreds place, the tens and ones place of the number are any of the following:

01, 04, 10, 13, 22, 31, 34, 40, 43

In such case 9 numbers of such 3-digit numbers are formed.

 

If the hundreds place digit is 3, then the sum of tens and ones place digits must be of the form 3z, where z is a whole number. After fixing 3 at hundreds place, the tens and ones place of the number are any of the following:

00, 03, 12, 21, 24, 30, 33, 42

In such case 8 numbers of such 3-digit numbers are formed.

 

Therefore, there are 16 + 9 + 8 = 33 numbers of 3-digit numbers formed by using the digits 0, 1, 2, 3 and 4.

Answered by  | 18th Sep, 2012, 12:57: PM

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