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Question 975898: what is the three digit number that satisfies the following condition.the tens digit is greater than the ones digit, the sum of the digits is 9 and if digits are reserved and resulting three-digit number is subtracted from the original number, the difference is 198
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let's call the three digits h, t, and o for hundreds, tens and ones places
The three numbers add to 9: h + t + o = 9
The number can be represented as 100h + 10t + o. Reversing gives 100o + 10t + h
If we subtract the reversed number from the original the difference is 198:
100h + 10t + o - 100o - 10t - h = 99h - 99o = 198 -> h - o = 2
In other words, the difference of the hundreds and ones digits is 2.
Since all three digits must add to 9, and we're told the tens digit is greater than the ones digit (so they can't be equal), the hundreds digit must be less than 7
But since the ones digit is two less than the hundreds digit,6 won't work because 6 + 4 = 10.
5 doesn't work either, since the tens digit > ones digit, and the ones digit must be 3 in this case, which would give 513.
432 works since they add to 9 and the tens digit > ones digit
351 also works for the same reasons
So the two possible numbers are 351 and 432
Check: 432 - 234 = 198 ; 351 - 153 = 198
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