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Question 1093195: A three digit no. is 12 less than 25 times the sum of it's digit. Reversing the digits increases the no. by 99. The tens digit is 1 more than the sum of hundreds and unit digits. Find the no.
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
Please if you want a solution using formal algebra, re-post saying so. This problem is much more easily solved with logical reasoning and a bit of trial and error.
The fact that reversing the digits increases the number by 99 means the hundreds digit is 1 less than the units digit. If you are not familiar with that fact, try a few examples.
172 --> 271; 271-172 = 99
314 --> 413; 413-314 = 99
etc., etc.
So we know the hundreds digit is 1 less than the units digit. Then given the information that the tens digit is 1 more than the sum of the hundreds and units digits, we only have a very few possibilities for the 3-digit number:
142
263
387
Now we only need to find which of these is 12 less than 25 the sum of its digits.
And we don't even have to try all three of those, because if the number is 12 less than a multiple of 25, the units digit must be either 3 or 8. The only one of our three possibilities that satisfies that condition is 263.
And checking the number 263, we find that indeed all the conditions of the problem are satisfied.
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