# Can you find the number?

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?

• You might enjoy it more if you refrain from using Wolfram Alpha. It’s not a typing puzzle. LOL
– asg
Jul 28 '20 at 15:30
• (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. Jul 28 '20 at 16:04
• 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.
– asg
Jul 28 '20 at 21:40
• @RossPresser For something more challenging, try this one: puzzling.stackexchange.com/q/100599/70406
– asg
Jul 31 '20 at 2:57
• You can find the answer using the three-equations three-unknowns system too Aug 6 '20 at 21:07

The number is

$$954$$,

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

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

$$4H+T=41$$, 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.

• Where are you getting that the number is three digits? Oct 31 at 15:12
• @ralphmerridew I guess I assumed it was implied from the wording of the question ... also the last restriction will put an upper bound on the size of the number, maybe that can be used to prove it must be three digits? Oct 31 at 15:47
• @Randal'Thor it's a good point actually, although probably not intended, perhaps it'll be good if we can prove that it needs to be 3 digits. Namely, it can't be 5 or more since 45*53 < 10000, and must be <2000 because 36*53 = 1908, and since 28*53 = 1484, it must be less than that. Since the last three digits is 899 or 954, it must be 3 digits. Nov 16 at 8:10

Alternate method using only the last clue: (N is number, S is digit sum)

$$N == 53*S$$
Since all numbers are congruent to their digit sum mod 9:
$$S === 53(S) (mod 9)$$
$$0 === 52*S (mod 9)$$
$$0 === 7*S (mod 9)$$
multiply both sides by 4
$$0 === 28*S (mod 9)$$
$$0 === S (mod 9)$$

So it's only necessary to check multiples of $$53*9$$. The sum of the digits in a number will always be nonnegative, so negative multiples are out. If the number is 0, then the third clue will involve 0/0. Therefore it's only necessary to check positive multiples of $$53*9$$.
$$53*9*1 == 477$$, digit sum is 18, not 9
$$53*9*2 == 954$$, digit sum is 18 MATCHES
$$53*9*3 == 1431$$, digit sum is 9, not 27
$$53*9*4 == 1908$$, digit sum is 18, not 36
$$53*9*5 == 2385$$, digit sum is 18, not 45
Note that the smallest number with a digit sum of 45 is 5 digits long, which is larger than $$53*9*5$$, and longer numbers will have even greater disparity, (exponential vs linear growth), so larger multiples of $$53*9$$ can't work.

All that remains is to check that $$954$$ satisfies the first two constraints: $$9+4-5 == 8$$
$$3*9+2*5-4 == 27+10-4 == 33$$

so $$954$$ is the only answer.