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The sum of the digits of a two digit number is 16 If the digits are reversed the new number is 18 less than the original number Find the original number?
In most simple problems like this, I prefer to solve it directly instead of using algebra. If the sum of the digits is 16, the digits can be 7 and 9, 8 and 8, or 9 and 7. So all we have to do is reverse the digits of 79, 88, and 97, to see which reversed number is 18 less than the original. But it should be pretty clear that we don't actually need to do that to all 3, because 97 is the only one that will get smaller by reversing the digits. I'm pretty sure by now that the answer is 97, but to be sure let's reverse the digits and check if the difference is 18: 97-79=18. Yep, 97's the answer!
This could also be solved using algebra, like they probably wanted you to do:
We'll use T for the tens digit and U for the units digit.
Given T+U=16 and 10T+U-18=10U+T, solve.
U=16-T
10T+16-T-18=10(16-T)+T 9T-2=160-9T
18T=162
T=9
U=16-9=7
So the original number was 97.
What else can I help you with?
A two digit number is 3 times the sum of each digit If 45 is added to number the digits are reversed Find the number.?
100000 2847239582
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A) If a number has two digits, then the sum of its digits is less than the value of the original two-digit number.
A two digit number is 3 times the sum of each digit If 45 is added to number the digits are reversed Find the number.?
100000 2847239582
