+0  
 
+5
1028
2
avatar+564 

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?

chilledz3non  Apr 25, 2016

Best Answer 

 #1
avatar+65 
+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
 #1
avatar+65 
+5
Best Answer

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
avatar
+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

Guest Apr 25, 2016

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