Question

While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number. Three times the sum of the digits of the original number is ten less than the number. What is the product of the digits of that number?

Please help how to obtain that number.

Answer 1

let the 10th digit be interchanged and in form 10x+y
therefore original sum
=> (a1+a2+...+a9) + 10x + y = S

but the sum obtained was::: (a1+a2+...+a9) + 10y + x = S' bcoz the digits were interchanged

"the sum of all the ten numbers increased by a value which was four less than that number"
therefore!!
S' - S = (10x+y) - 4
the a1, a2...a9 gets cancelled out & what is left is
=> 9(y-x)=10x+y-4 => 8y-19x=(-4)-----------------------------...

"three times the sum of digits of that original number is ten less than the number"
therefore
=> 3(x+y) = 10x+y - 10 => 2y-7x = -10-------------------------------------...


by (1) & (2), the y can be cancelled out by multiplying (2) with -4 & adding (1) & (2)

the result we get is
=> 9x = 36 => x=4

therefore by (2)
2y -7.4= -10 => y=9

therefore x.y = 36