Question

Find the number of three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.

Answer 1

Answer:

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Answer 2

There are 2 such 3-digit numbers that on dividing by 11 are equal to the sum of the squares of the digit of the initial number.

Given

three-digit numbers when dividing the number by 11, the sum of the squares of the digits of the initial number

To Find

Number of such numbers present

Solution

Let the 3-digit number be 100a + 10b + c

where, a, b, and c are the three digits. and a>0

According to the problem

(100a + 10b + c)/11 = a² + b² + c²

or, 100a + 10b + c = 11a² + 11b² + 11c²

Since these numbers when divided by 11 results in whole numbers, they must be divisible by 11

For a 3-digit number to be divisible by 11

1. a + c - b = 0 or, b = a + c or, c = b - a
2. a + c = 11

Now scanning all such numbers from 100 to 999 we get 2 such numbers

550 and 803

we see that

550/11

= 50

also, 5² + 5² + 0²

= 25 + 25

= 50

Therefore, this satisfies the condition.

Also,

803/11

= 73

8² + 0² + 3²

= 64 + 9

= 73

Therefore, this too satisfies the condition

Therefore, there are 2 such 3-digit numbers that on dividing by 11 are equal to the sum of the squares of the digit of the initial number.

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