the sum of the digits of a 2 digit number is 11. if the number obtained by reversing the order of digits is 9 less than the original number , find the orginal number .
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Step-by-step explanation:
Let T be the tens digit and U be the units digit. Then the value of the original number is 10T+U. The new number (with reversed digits) will have a value of 10U+T. We are told that the new number is 9 less than the original number:
10U+T=10T+U−9
9U−9T=−9
U−T=−1
We are also told that the sum of the digits of the original number (and also the new number, since the digits are the same) is 11:
U+T=11
Combining these two equations by adding, we get:
2U=10 or U=5
Since the two digits add up to 11, that leaves T=6 , and the original number was 10(6)+(5)=65
Now, here’s a shortcut for this kind of problem. If the difference between the new and old numbers is 9, the digits will differ by 1. If the difference is 18, the digits will differ by 2. If the difference is 27, the digits will differ by 3, and so on. It’s usually pretty easy to solve these in your head once you know how to find the difference between the digits. In the original problem, you might think: Ah, okay. They add to 11 and differ by 1. Must be 6 and 5. So the two numbers are 65 and 56. Now, which number was bigger? The problem says the new number is LESS, so the original number must have been 65.