Question

What is the sum of digits of a 2-digit number which is 40 less than the square of the product of its digits?

Answer 1

29.......

if it is correct please follow me and Mark it as brainy please.......

Answer 2

ANSWER:  

The sum of digits of a 2-digit number which is 40 less than the square of the product of its digits is 6

SOLUTION:

Given, 2-digit number is 40 less than the square of the product of its digits.

Now, let us consider some of the square terms  

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 125.  

We need not consider above 125, because after subtracting 40 the result should be a two digit number.

Now, we can also discard 1, 4, 9, 16, 25, 36, 49. As by subtracting 40 from these numbers we don’t get a two digit number.

Now remaining squares are 64, 81, 100, 125

After subtracting 40 above terms becomes as 24, 41, 60, 85

Now product of digits is 8, 4, 0, 40. After squaring these terms we should get the previous square terms.

$8^{2}=64,4^{2} \neq 81,0^{2} \neq 100,40^{2} \neq 125$

So, the only number satisfying our conditions is 24.

Hence, sum of digits of the two digit number is 2 + 4 = 6.