Question

A two-digit number ab is added to the number formed by reversing the original digits. If their sum is
divisible by 11, 9 and 2. Find the number of pairs of (a, b).

Solution : The original number is ab.
Value of the number 10 a + b.
Value of the number by reversing the digits = 10 b + a
Sum of the two numbers 10 a + b + 10 b + a = 11 a +11 b = 11 (a + b)
Given that sum is divisible by 11, 9, 2, i.e., (a + b) must be divisible by both 9 and 2.
a + b = 8
It is possible only when a = b = 9.
Hence, original number is 99.​

someone pls explain thel last 2 points as to how did we come at a+b = 8, pls?

Answer 1

Answer:

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Step-by-step explanation:

Given units digit is x and tens digit is y

Hence the two digit number = 10y + x

Number obtained by reversing the digits = 10x + y

Given that sum of a two digit number and the number obtained by reversing the order of its digits is 121.

Then (10y+x)+(10x+y)=121

⇒10y+x+10x+y=121

⇒11x+11y=121

⇒x+y=11

Thus the required linear equation is x + y = 11.