Question 127974This question is from textbook Structure and Method Book 1
: When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?
This question is from textbook Structure and Method Book 1
Answer by ilana(307)   (Show Source):
You can put this solution on YOUR website! Let's say the original number is ab, so a is the tens digit and b is the ones digit. That means the reversed number is ba, where b is tens and a is ones. Look at the number 35. 3 is the tens and 5 is the ones, so the total is 3(10)+5(1)=30+5=35. So ba must actually be 10b+a. We know this is 9 more than the original, so 10b+a=10a+b+9. If the digits add to 11, a+b=11.
Now we have a system of 2 equations: 10b+a=10a+b+9 and a+b=11. Using the first, 10b+a=10a+b+9 becomes 9b=9a+9, divide by 9 and get b=a+1. Plug that b into a+b=11 and get a+(a+1)=11, so 2a+1=11, so 2a=10, and finally, a=5. If a=5, since a+b=11, b=6. So the original number was 56. Check if this is right: 65-56=9, and 5+6=11. Nicely done!:)
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