Numerical CAT

Number Theory Level 2

Let $\overline{CAT}$ be a three digit number with $C$ being the hundreds digit, $A$ being the tens digit and $T$ being the ones digit.

$\overline{CAT}=(C+A+T)\times C\times A\times T$ .

Given that $T$ is even, what is the value of $\overline{CAT}$ ?

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The answer is 144.

1 solution

Saya Suka
Jan 8, 2021

First we have to acknowledge that none of C nor A nor T can neither be 0 nor 5. Since we're told that T is even, any factor of 5 between C or A will make T a 0 and that effectively rendered each of C, A & T to also become 0.

100C + 10A + T
= (C + A + T)CAT
= (99C + 9A) + (C + A + T)
9(11C + A) = (C + A + T)(CAT - 1)
LHS has a factor of 9, so RHS must have the same.
Either CAT = 1 mod 9, or
C + A + T = 0 mod 9, or
both {CAT = 1 mod 3 & (C + A + T) = 0 mod 3}
Considering T is even, then we know that
CAT = {10, 28, 46, 64, 82}
Reject both 46 and 82 for the two having prime factors > 7. We're left with candidates of 100C + 10A + T among {152, 512, 272, 722, 174, 714, 444, 248, 284, 428, 482, 824, 842, 188, 818}.
The first permutation has a sum of 1+2+5 = 8 with no multiplier effect from (10-1)/9, too low to match up to LHS's 11C + A. The second one has a sum of 2+2+7 = 11, but even with multiplier of 3, it can't match up to LHS when we know we cannot have A = 0. The third has a sum of 1+4+7 = 12 with multiplier 3, but 11C + A ≠ 36 without any 3 for C (4 is too high and 1 too low). The fourth has a sum of 4+4+4 = 12 with multiplier 7, but 11C + A ≠ 84 without any 7 for C (4 is just too low). The fifth has a sum of 2+4+8 = 14 with multiplier 7, but 11C + A ≠ 98 since 11x8 would need an extra 10 to reach 98, but our A is a one digit number less than 10. The sixth and last one has a sum of 1+8+8 = 17 with multiplier also 7, but 11C + A ≠ 119 since C would have to be a two digit number of 10 to match up the RHS's value.

Now that we have exhausted all and any possibility of CAT = 1 mod 9, we'll move on to C + A + T = 0 mod 9.
Then C + A + T = 9 or 18 (but not 27 for the mention of T's evenness).
Our list would look like this :
Permutations of {(1,2,6) , (1,4,4) , (2,3,4) , (1,8,9) , (2,7,9) , (2,8,8) , (3,6,9) , (3,7,8) , (4,6,8) , (4,7,7) , (6,6,6)}.
For the first, they multiply into 12, and subtracting 1 from it, LHS's 11C + A cannot match an 11 without A = 0, a contradiction to our first paragraph. But the second turns out nicely, 1x4x4 -1 = 15 for RHS, and RHS = 11x1+4 = 15. Trying it out,
1x4x4x(1+4+4) = 16x9 = 144.
Yay, finally the answer of
100C + 10A + T = 144

Numerical CAT

Image Credit: Wikimedia Cat silhouette

The answer is 144.

1 solution

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