SOLUTION: 2) The sum of the digits of a two- digit counting number is 6. When the digits are reversed, the number is 18 greater than the original number. What was the original number?

Question 1197732: 2) The sum of the digits of a two- digit counting number is 6. When the digits are reversed, the number is 18 greater than the original number. What was the original number?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52818)   (Show Source): You can put this solution on YOUR website!
.
The sum of the digits of a two- digit counting number is 6.
When the digits are reversed, the number is 18 greater than the original number.
What was the original number?
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Let the digits of the original number be "a" and "b", reading from left to right. Then the number is 10a+b, and we are given thast a + b = 6. (1) The reversed number in written form is "ba" : its value is 10b+a. So, we have second equation (10b+a) - (10a+b) = 18, or 9b - 9a = 18 9(b-a) = 18 b - a = 2. (2) Thus you have two equations, (1) and (2), to find "a" and "b". To do it, express b = 6-a from (1) and substitute into (2). You will get (6-a) - a = 2 6 - 2a = 2 6 - 2 = 2a 4 = 2a a = 4/2 = 2. Thus you found a= 2, b= 6-a = 6-2= 4. Hence, the original number is 24. ANSWER

Solved, with full explanations.

Answer by josgarithmetic(39621)   (Show Source): You can put this solution on YOUR website!
Number,

May be able to see without making steps on paper,
u=4 and t=2;

The two-digit number, 24.
Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!

The responses from the other tutors show good formal algebraic solutions.

If a formal algebraic solution is not required and the speed of solving the problem is important (as on a timed competitive test), then this kind of problem can be solved quickly using this fact:

The difference between a 2-digit number and the 2-digit number with the digits reversed is 9 times the difference of the digits.

It is easy to show this. If t and u are the tens and units digits of the original number, then the original number is 10t+u and the number with the digits reversed is 10u+t. The difference between the two 2-digit numbers is

(10t+u)-(10u+t)=9t-9u=9(t-u)

So in this problem, with a difference of 18 between the two 2-digit numbers, we know the sum of the digits is 6 and the difference is 2. Logical reasoning and/or simple arithmetic or algebra shows us the two digits are 2 and 4.

Then, since the number with the digits reversed is larger, the original number is 24.

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