There's a number with the following characteristics:

  • The hundreds digit plus the units digit minus the tens digit equals 8.
  • 3 times the hundreds digit plus 2 times the tens digit minus the units digit equals 33.
  • The number divided by the sum of its digits equals 53.

What's the number?

  • $\begingroup$ You might enjoy it more if you refrain from using Wolfram Alpha. It’s not a typing puzzle. LOL $\endgroup$
    – asg
    Jul 28 '20 at 15:30
  • $\begingroup$ (deleted my spoilery wolframalpha comment.) My point is that there isn't much puzzle to this. It's a problem in translating words to algebra, suitable for homework, not this SE. But I won't argue; at least 7 people disagree with me. $\endgroup$ Jul 28 '20 at 16:04
  • 1
    $\begingroup$ I agree, it wasn't much of a challenge. That was just my first run to get my feet wet. It only took a couple of minutes to create and verify, so I didn't expect it to take too long to solve. I trust my next one will pose more of a challenge for you. BTW, my years tutoring math have convinced me that translating words into equations is a significant puzzle for many. The fact that some are very good at it does not diminish its difficulty. $\endgroup$
    – asg
    Jul 28 '20 at 21:40
  • $\begingroup$ @RossPresser For something more challenging, try this one: puzzling.stackexchange.com/q/100599/70406 $\endgroup$
    – asg
    Jul 31 '20 at 2:57
  • $\begingroup$ You can find the answer using the three-equations three-unknowns system too $\endgroup$
    – aminabzz
    Aug 6 '20 at 21:07

The number is


because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively:

$H+U-T=8,\quad 3H+2T-U=33,\quad 100H+10T+U=53H+53T+53U$

Adding the first two gives

$4H+T=41$, therefore $T=41-4H$. Since both $T$ and $H$ must be numbers from $0$ to $9$, the only possibilities are $H=8,T=9$ and $H=9,T=5$.

Also from the first equation,

$U=8+T-H=8+(41-4H)-H=49-5H$. Here the only possibilities are $H=8,U=9$ and $H=9,U=4$.

So the number must be

either $899$ or $954$. Only one of these is a multiple of $53$, namely $954=53\times18=53\times(9+5+4)$, so the problem is solved.


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