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 reversed and resulting three-digit number is subtracted from the original
number, the difference is 198
Let the number be ABC and its reverse be CBA,
Then the subtraction is:
ABC
-CBA
---
198
A must be bigger than C so that the top number will be bigger than the bottom
number. So we will need to borrow 1 from the B at the start of the subtraction.
That will reduce the B in the top middle digit to 1 less than the B just below
it.
Therefore we will also need to borrow 1 from the A as well.
If we had not needed to borrow 1 from the A, then A-C would have been 1 to
get the 1 in 198. But since we must borrow 1 from A, A-C must be 2
instead, so that borrowing 1 from A would give the 1 in 198.
A-C cannot be as high as 6-4 because all three digits must have sum 9,
and 6 and 4 alone have sum 10. So we try lower digits.
We try A-C = 5-3, making the number 513 since the digits must have sum 9.
But that won't do because the tens digit is less than the ones digit.
We try A-C = 4-2. Then the number would be 432 since the digits must have sum 9.
That's one solution because the tens digit is greater than the ones digit, and
432-234=198.
We try A-C = 3-1. Then the number would be 351 since the digits must have sum 9.
That's another solution because the tens digit is greater than the ones digit,
and 351-153 = 198.
So there are TWO solutions, 432 and 351
432 351
-234 -153
---- ----
198 198
Edwin