Question

The sum of a two digit-digit number and the number obtained by interchanging the digits is always divisible by which of the following numbers? *​

Answer 1

To find:

✠ The sum of a 2-digit number and the number obtained by interchanging the digits is always divisible by which number?

Solution:

Choose any 2-digit number xy = 10x + y,

So the 2-digit number is in the form 10x + y.

The number obtained by interchanging i.e, reversing the digits yx = 10y + x.

On adding the two numbers, i.e, adding the reversed number to the original number.

⟼ ( 10x + y ) + ( 10y + x )

⟼ 11x + 11y

Taking out the common factor in both the terms, we have:

⟼ 11 ( x + y )

So, the sum is always a multiple of 11.

Now, divide the sum by 11.

⟼ 11 ( x + y )/11

⟼ x + y

Here, we get no remainder and we can easily observe, that the quotient is the sum of digits of original 2-digit number xy.

∴ The sum of a 2-digit number and the number obtained by interchanging the digits is always divisible by 11.

_______________________________

Answer 2

Answer:

The sum of a 2-digit number and the number obtained by interchanging its digits is always divisible by 11. The difference between a 2-digit number and the number obtained by interchanging its digits is always divisible by 9.

Step-by-step explanation:

Hope it helps..