A two-digit number is 6 more than the product of 7 and the sum of its digits. The number with the digits reversed is two more than the product of 3 and the sum of its digits. What is the number?

Answer 1

it can be done by guessing, but here is the algebraic solution

x is the Tens place...

y is the Ones place

The two digit number is 10x+y...

the sum of the digits is x+y...

The two digit number is 6 more than the product of 7 and the sum of it's digits:

10x+y = 6 + 7(x+y) is the first equation

Reversing the digits, the number with the digits reversed is 2 more than the product

of 3 and the sum of the digits:

10y + x = 2 + 3(x+y) is the 2nd equation

1st equation:

10x+y = 6 + 7(x+y)

10x + y = 6 + 7x + 7y <--- distributive

3x - 6y = 6 <--- moves all the vars to left side and combines

x - 2y = 2 <--- divides by 3

x= 2y + 2 <--- solves for x

2nd equation:

10y + x = 2 + 3(x+y)

10y + x = 2 + 3x + 3y <--- distributive

7y - 2x = 2 <--- moves all vars to left side and combines

Finally substitutes the first equation into the 2nd equation:

7y - 2(2y+2) = 2

7y - 4y - 4 = 2

3y - 4 = 2

3y = 6

y = 2

then x = 2y+2 = 2(2)+2 = 4+2=6

the number is 62....

note that 62 = 6 + 56 = 6 * 7*8 = 6 + 7*(6+2) <-- 6 more than 7 times the sum of the digits

and

26 = 2 + 3(2+6), so yes it checks

the number is 62