Divisibility test

Let $a$ and $b$ be the digits in ten's and one's place respectively of the original positive number.

Now, we can express the original number as $10a+b$ .

When the digits are reversed, the digits in ten's place and one's place become $b$ and $a$ respectively.

Then, the new number can be expressed as $10b+a$ .

Sum of the two numbers $=10a+b+10b+a =11a+11b = \boxed{11(a+b)}$ [where $a,b$ are (+ve) nos.]

Since the sum is a multiple of $11$ , so the sum will always be divisible by $\boxed{11}$ .

Let the number be 10x+y, on reversing it becomes 10y+x. Adding both, you get 11(x+y)

Thus it is divisible by 11

Let $x$ be the tens digit and $y$ be the units digit of the original number respectively .Then the original number is $10x+y$ because all 2 digit numbers can be written this way Example: $23=10(2)+3,57=10(5)+7\; etc$ Anyways,if we reverse the digits of the number the new number is $10y+x$ .Let`s add them together: $(10x+y)+(10y+x)$ $10x+x+10y+y$ $11x+11y$ Observe that $11x+11y=11(x+y)=11\times\;some\;number$ So the sum will always be divisible by $\boxed{11}$

The number will always be divisible by 11 because the digits will all be the same. That is because you are adding the same digits in the ones and tens place.

Two digit no will have 10multiplied by x and unit number will be y say. reverse number ill be 10 Y +X when we add up sum will be 11(X+Y) so it will be divisible by 11 Ans.

K.K.GARG,India

If you reverse and add the two numbers, both digits of the sum are the same number. All two digit numbers that are like this are divisible by 11. i.e. 11, 22, 33, 44, 55, ...

Let a number be 12 and inverse one be 21.Adding them we get 33, which is divisible by 11. (SOLVED)

take two positive integer it will become two digit number if any two digit number is reversed and added to the original it will divisible by 11 as given by Prasun Biswas

$\overline{ab}$ is a two-digit number.

We can replace him with $10a + b$ .

So , $\overline{ba} = 10b + a$

Adding: $(10a + b) + (10b + a) = 11 (a + b)$

So, the sum is always divisible by $11$ .

It is $(10a+b)+(10b+a) = 11a+11b$

$=11(a+b)$

For example, we take 72.

Reverse its digits and we get 27.

The sum of both numbers is 99, which existed in multiplication of 11. Thus, it will be always can be divided by 11.

Digit at 10th number will become digit at oneth number : add both we get 11 times the number so it is always divisible by 11

For example take any number like 34 reverse it we get 43.If we add both we will get 77.It will be divided by11