Transcript Numbers and Algebra

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Think of a 3-digit number such that the first
and the last digit differ by 2 or more.
E.g. 246
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Reverse the digits in the 3-digit number
Subtract the smaller 3-digit number from the
larger one.
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Reverse the digit in the answer and add it to
the original answer.
Keep the answer to yourself and don’t let
others get to know it.
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Don’t tell me your answer because I can guess
yours.
Why?
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The number is 123.
Reverse it becomes 321.
Then 321 – 123 = 198
Now, reverse 198 to become 891.
Then 891 + 198 = 1089.
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This is only true to a particular number…
How to say this is true in general?
We need the
help from
ALGEBRA!
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In general, the 3-digit number is 100a + 10b + c.
Reverse it to become 100c + 10b + a.
Subtract one number from the other
(100a + 10b + c) – (100c +10b + a)
= 100a – a + 10b – 10b + c – 100c
= 99a – 99c
= 99 (a – c)
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Since we need to keep the 99 (a – c) to be 3digit, a – c ≥ 2.
Possible values are 2, 3, … ,9
Possible values for 99 (a – c) are 198, 297, 396,
495, 594, 693, 792, 891.
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Check yourself if you have one of those
numbers before the addition of its reverse…
Addition of those numbers and its reverse will
always be equal to 1089.
198  891  1089
594  495  1089
297  792  1089
693  396  1089
396  693  1089
792  297  1089
495  594  1089
891  198  1089