The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 9 less than three times the original number. What is the original number?

Thank you!

Question

The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 9 less than three times the original number. What is the original number?

Thank you!

Isaiah Anthony · Accepted Answer

Number is #27#.

Let the unit digit be #x# and tens digit be #y# then #x+y=9# ........................(1) and number is #x+10y# On reversing the digits it will become #10x+y# As #10x+y# is #9# less than three times #x+10y#, we have #10x+y=3(x+10y)-9# or #10x+y=3x+30y-9# or #7x-29y=-9# ........................(2) Multiplying (1) by #29# and adding to (2), we get #36x=9xx29-9=9xx28# or #x=(9xx28)/36=7# and hence #y=9-7=2# and number is #27#.

Nathan Axel · Answer

undefined Let the original two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit. According to the given conditions:

1. x + y = 9 (sum of the digits is 9)
2. 10y + x = 3(10x + y) - 9 (reversed number is 9 less than three times the original number)

Solving these equations simultaneously:

From equation 1, we have: x = 9 - y

Substituting x = 9 - y into equation 2:

10y + (9 - y) = 30x + 3y - 9
9y + 9 = 30(9 - y) + 3y - 9
9y + 9 = 270 - 30y + 3y - 9
9y + 9 = 270 - 27y
36y = 261
y = 7

Substitute y = 7 into equation 1:

x + 7 = 9
x = 2

Therefore, the original two-digit number is 27.