Answer:
731
Explanation for problem:
"The sum of the digits of a three-digit number is 11."
x + y + z = 11
The three digit number = 100x + 10y + z
The reversed number = 100z + 10y + x
" If the digits are reversed, the new number is 46 more than five times the old number."
100z + 10y + x = 5(100x + 10y + z) + 46
100z + 10y + x = 500x + 50y + 5z + 46
combine on the right
0 = 500x - x + 50y - 10y + 5z - 100z + 46
499x + 40y - 95z = -46
"the hundreds digit plus twice the tens digit is equal to the units digit,"
x + 2y = z
x + 2y - z = 0
Three equations, 3 unknowns
x + y + z = 11
x +2y - z = 0
-----------------Addition eliminates z
2x + 3y = 11
From the 2nd equation statement, we know that the 1st original digit has to be 1
2(1) + 3y = 11
3y = 11 - 2
3y = 9
y = 3 is the 2nd digit
then
1 + 3 + z = 11
z = 11 - 4
z = 7
137 is the original number
Check solution in the 2nd statement
If the digits are reversed, the new number is 46 more than five times the old number."
731 = 5(137) + 46
731 = 685 + 46
