Question

ab and ba are 2 two-digit numbers such that ab + ba = cac. what is a+b+c ? (assume c is not 0 )

Answer 1

So,

ab + ba = cac

This situation can be written numerically as,

10a + b + 10 b + a = 100c + 10a + c

11a + 11b = 10a + 101c

a + 11b = 101c

Now, a and b are digits,

Hence they are 1 digit numbers

So,

The value of c must be definetly 1 ,

Reason :

Let us take a and b as 2 digit numbers,

10 + 110 = 120

Now,

120 = 101 c

So,

c can be only 1 because c= 2 will make a and b 2 digit numbers

So,

a + 11b = 101

So,

The only combination which satisfies this is

(2, 9)

So,

a= 2 , b= 9 ,c =1

So,

a + b +c = 2 + 9 + 1 = 12

So,

The number is 12

Answer 2

Acc to the question, there are 2 two digit numbers and 1 three digit number.

As per the data given, the numbers will be assumed as:

ab=(10a + b), ba=(10b + a) and cac=(100c + 10a + c).

Also, it is given that (10a + b) + (10x + a) = (100c + 10a + c)

or, We can write. ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11(a+b)

Also, ab + ba = cac

While adding two 2 digit numbers to get a 3 digit number, the maximum carry possible is 1. Without the carry it is not possible to get a 3 digit number. Therefore it is evident that c is equal to 1.

ab + ba = 1a1

you can see b+a = 1 and a+b gives a with carry 1.

So, definitely b + a = 11.

Now the carry 1 + a + b in tens place we get, 10 + a

on simplifying we get, b = 9.

Now the scenario is as such, a9 + 9a = 1a1

9 + a = 1 (actually its 11 with carry 1)

so, a = 2.

Now 29 + 92 = 121 [Ans]