+0

# What is the original number?

+5
2636
2

The sum of the digits of a two-digit counting number is 10. If the digits are reversed, the new number is one less than twice the original number. What was the original number?

Apr 25, 2016

#1
+5

I think it's 91.

9+1=10

10x2=20

We reverse 91 to 19

19 is one less than 20.

I am not sure though.

Apr 25, 2016
edited by Lunar  Apr 25, 2016

#1
+5

I think it's 91.

9+1=10

10x2=20

We reverse 91 to 19

19 is one less than 20.

I am not sure though.

Lunar Apr 25, 2016
edited by Lunar  Apr 25, 2016
#2
+5

A + B=10

2[10A + B] - 1=10B + A

So the original number is=37

Solve the following system:
{A+B = 10 |     (equation 1)
2 (10 A+B)-1 = A+10 B |     (equation 2)
Express the system in standard form:
{A+B = 10 |     (equation 1)
19 A-8 B = 1 |     (equation 2)
Swap equation 1 with equation 2:
{19 A-8 B = 1 |     (equation 1)
A+B = 10 |     (equation 2)
Subtract 1/19 × (equation 1) from equation 2:
{19 A-8 B = 1 |     (equation 1)
0 A+(27 B)/19 = 189/19 |     (equation 2)
Multiply equation 2 by 19/27:
{19 A-8 B = 1 |     (equation 1)
0 A+B = 7 |     (equation 2)
Add 8 × (equation 2) to equation 1:
{19 A+0 B = 57 |     (equation 1)
0 A+B = 7 |     (equation 2)
Divide equation 1 by 19:
{A+0 B = 3 |     (equation 1)
0 A+B = 7 |     (equation 2)
Collect results:
Answer: |  A = 3             and                  B = 7

Apr 25, 2016