The sum of the digits of a three-digit number is 11. if the digits are reversed, the new number is 46 more than five times the old number. if the hundreds digit plus twice the tens digit is equal to the units digit, then what is the number?

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

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