Question

If a two digit natural number is added to a number made by putting the digits of the original number in reverse order, the sum will always be divisible by

Answer 1

Answer:

Please redefined the question

Answer 2

The given question is If a two-digit natural number is added to a number made by putting the digits of the original number in reverse order, the sum will always be divisible by.

The solution to the given question is we have to find the number which divides the addition of the number in the original order and then it's the reverse order.

let the original order of the number be ab and the reverse order of it be ba Then their sum will be divisible by 11

let us prove this statement by using some examples such as

34, 89, 14.

the reverse order of the above numbers are

$34 \: \: \: 89 \: \: \: 14 \: is \: 43 \: \: 98 \: \: 41respectively$

adding the original and reverse order number gives the final answer as

$34 + 43 = 77 \\ 89 + 98 = 187 \\ 14 + 41 = 55$

The answers to the addition of numbers are 77,187,55.

These numbers are found to be divisible by the digit 11.

if we can take some more examples also, we will get the same numbers that are divisible by 11.

Hence the answer to the given question is 11.

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