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Numbers and Algebra. Step 1. Think of a 3-digit number such that the first and the last digit differ by 2 or more. E.g. 246. Step 2. Reverse the digits in the 3-digit number Subtract the smaller 3-digit number from the larger one. Step 3.
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Step 1 • Think of a 3-digit number such that the first and the last digit differ by 2 or more. • E.g. 246
Step 2 • Reverse the digits in the 3-digit number • Subtract the smaller 3-digit number from the larger one.
Step 3 • 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.
Let’s say… • The number is 123. • Reverse it becomes 321. • Then 321 – 123 = 198 • Now, reverse 198 to become 891. • Then 891 + 198 = 1089.
But… • This is only true to a particular number… • How to say this is true in general?
Let Algebra do the talking… • 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)
Let Algebra do the talking… • Since we need to keep the 99 (a – c) to be 3-digit, a – c ≥ 2. • Possible values are 2, 3, … ,9 • Possible values for 99 (a – c) are 198, 297, 396, 495, 594, 693, 792, 891.
Let Algebra do the talking… • 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.
