I’m revising some fundamentals (place value, digit sums, and those little divisibility tricks), and I ran into a number puzzle that’s got me second-guessing myself. The puzzle: Find a three-digit number where the sum of its digits is 15, the tens digit is 3 more than the ones digit, and if you reverse the digits you get a number that’s exactly 297 more than the original.
I keep wanting to just list possibilities, but I’m trying to build better reasoning muscle. What’s a clean way to set this up and spot the key relationships without brute force? Not looking for the answer-just how you’d think about it step by step, or a nudge in the right direction.
Any help appreciated!
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3 Responses
I’d set it up like a little place-value story: let the hundreds, tens, ones be a, b, c. Reversing abc to cba is like swapping the hundreds and ones suitcases; the tens just ride along. I always half-remember the effect as “about 100(c−a)”, but when you write it out, the tens cancel exactly and you get cba − abc = 99(c − a). That 297 clue practically shouts 99 × 3, so c − a = 3. You’ve also got “tens is 3 more than ones,” so b = c + 3, and the digit-sum a + b + c = 15. That’s the whole system without listing anything: c = a + 3, b = c + 3, and a + b + c = 15.
From there it’s just a quick substitution: b = a + 6 and c = a + 3, so a + (a + 6) + (a + 3) = 15 → 3a + 9 = 15 → a = 2, then b = 8 and c = 5. So the number looks like 285, and its reverse 582 is indeed 297 more. I’m 95% sure there isn’t some sneaky edge case I’m missing (I briefly worried about “carries,” but that matters for addition, not for this difference), and a nice side-check is the digit sum 15, which makes the number divisible by 3-though not by 9; I always almost mix those rules. If you like this style of setup, this short walkthrough on digit word problems is handy: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations-word-problems/x2f8bb11595b61c86:digit-word-prob/v/digit-problem and a quick refresher on the 3-and-9 divisibility trick: https://www.khanacademy.org/math/arithmetic/arith-review-multiplication-division/arith-review-divisibility-tests/v/divisibility-rules-3-and-9.
Let the number be 100a + 10b + c; reversing gives 100c + 10b + a = (100a + 10b + c) + 297, so 99(c − a) = 297 ⇒ c − a = 3, and with b = c + 3 and a + b + c = 15 you get 3a + 9 = 15, then solve for a and back out c and b.
Analogy: like three meshed gears-reverse fixes the gap between first and last, “+3” sets the middle, and the digit sum locks everything in place.
Think algebraically: let the digits be a (hundreds), b (tens), c (ones); then a+b+c=15, b=c+3, and “reversed is 297 more” gives 100c+10b+a=100a+10b+c+297, which tidily simplifies to c−a=3.
So c=a+3 and b=a+6, hence 3a+9=15 → a=2, giving (a,b,c)=(2,8,5), i.e., 285-quick check: 582−285=297.