SOLUTION: The sum of the two digits of a positive integer is 12. When the digits were reversed, the new number was 54 greater than the original. What is the product of the digits of the orig
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Question 1156074: The sum of the two digits of a positive integer is 12. When the digits were reversed, the new number was 54 greater than the original. What is the product of the digits of the original number minus the original number?
Using formal algebra (way too much work; but you should understand how to solve the problem this way....)
t = tens digit
u = units digit
t+u=12 [the sum of the digits is 12]
10t+u = original number
10u+t = number with digits reversed
10u+t = 10t+u+54 [the number with the digits reversed is 54 more than the original number]
9u-9t = 54
u-t = 6
t+u=12 and u-t=6 --> u=9, t=3
The original number was 39.
Then answer the question that is asked....
If just getting the answer quickly is the objective (formal algebra not required), then a bit of simple arithmetic can solve the problem in a few seconds.
One shortcut would be to use the fact that the difference between a 2-digit number and the number with the digits reversed is 9 times the difference of the digits. (That can be seen in the formal algebra above).
That immediately gets you to t+u=12 and u-t=6, leading to 39 being the original number.
Even faster than that is simply to look at 2-digit numbers where the sum of the digits is 12. The number with the digits reversed is bigger than the original number, so the tens digit is the smaller of the two digits. Then you just need to look at the numbers 57, 48, and 39 and see which one will get bigger by 54 when you reverse the digits.
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