A number theory cryptogram

In this solution, the equivalence $\equiv$ means modulo 10 unless otherwise stated.

Last digit: $T^2 \equiv T$ . This requires $T = 0, 1, 5, 6$ .

Cross product of last two digits: $2AT \equiv A +\$ tens digit of $T^2$ . Split out the cases:

• If $T = 0$ , the product should end in two zeroes, but this is impossible since $A \not= T$ .

• If $T = 1$ , the cross product $2A \equiv A$ . This means $A = 0$ . Now $CAT^2 = C01^2 \equiv 200C + 1$ modulo 10000, requiring $B = 0$ . But this is impossible since $A \not= B$ .

• If $T = 5$ , the second last digit must be $A = 2$ . We get $CAT^2 = C25^2 \equiv 625 + 5000C$ modulo 10000. This implies $BC \equiv 50C + 6$ , which immediately gives $C = 6$ and $B = 0$ . But $B \not= 0$ because it occurs as initial digit of the product.

• That leaves $T = 6$ and $12A + 3 \equiv A$ , which implies $A = 7$ . The equation $C76^2 = BOBC76$ translates into $5776 + 15200\ C + 10000\ C^2 = BOBC76 \\ 57 + 152C + 100C^2 = BOBC \\7 + 2C \equiv C,$ from which $C = 3$ . Check: $376 \times 376 = 141376$ is indeed a solution.

Thus $B + O + C + A + T = 1 + 4 + 3 + 7 + 6 = \boxed{21}$ .

Let $x$ be the 3-digit integer, $\overline{CAT}$ , then by looking at the long multiplcation, we can see that the last 3 digits remains the same after the multiplication. That is $x^2 \equiv x \pmod{1000}$ .

Or equivalently $x(x-1) \equiv 0 \pmod{1000}$ . Since $x$ and $x-1$ are coprime, and $1000 = 125\times8$ . Then one of $x,x-1$ has to be a multiple of 125, while the other is a multiple of 8.

We have either

Case 1: $\begin{cases} x \equiv 0 \pmod{125} \\ x \equiv 1 \pmod 8\end{cases}$ . Solving via Chinese remainder theorem (CRT) gives $x=625$ only. But $x^2 = 390625 =\overline{BOBCAT}$ , so $B = 3$ and $B= 0$ at the same time, which is impossible.

Case 2: $\begin{cases} x \equiv 1 \pmod{125} \\ x \equiv 0 \pmod 8\end{cases}$ . Solving via CRT again yields $x = 376\Rightarrow x^2 = 141376 = \overline{BOBCAT}$ .

So the only possible scenario is Case 2 with $B = 1, O = 4, C = 3, A = 7, T = 6 \Rightarrow B+O+C+A+T=\boxed{21}$ .

heres an approach, we know 3 modular equations that can be easily solved: $\begin{cases} T^2 \equiv T \mod{10}\\ (10A+T)^2 \equiv 10A+T \mod{100}\\ (100C+10A+T)^2 \equiv 100C+10A+T \mod{100}\\ \end{cases}$ it easy to see the first equation has only 4 solution less than 10 $T=(0,1,5,6)$ . we can eliminate $T=0$ off the bat since it implies $A=0$ which contradicts the constraint. we can see from the second equation: $(10A+T)(10A+T-1)\equiv 0\mod{100} \implies 10(2T-1)A+T^2-T\equiv 0 \mod{100}$ since {10|T^2-T) we can see 10 divides both of the quntits been added, so when divided by tten, another 10 must also divide the result; i.e $(2T-1)A+\dfrac{T^2-T}{10}\equiv 0 \mod{10}$ we can now try plugging in the values of T we know to get solutions: $(A,T)= (0,1),(2,5),(7,6)$ let $n=10A+T$ , then in the 3rd eqn: $(100C+n)(100C+n-1)\equiv 0 \mod{1000} \to 100(2n-1)C+n^2-n \equiv 0 \mod{1000}$ by using a similar logic used before, $(2n-1)C+\dfrac{n^2-n}{100} \equiv 0 \mod{10}$ and by plugging in all the values of n, we have $(C,A,T)= (0,0,1), (6,2,5),(3,7,6)$ we can test these solutions to get the only solution satisfying the constraints is: $376 \times 376=141376$ $1+4+3+7+6=\boxed{21}$

Answer is 376*376=141376

Solving the units place we have T=0,1,5,6

Solving tens place , we have AT=00,01,25,76

Solving 100 place , we have CAT=000,001,625,376.

Actual squaring shows that only 376 satisfies all the conditions.

To solve, I mean to say that : $x^2=10k+x$ For units place. (Ideally solved by finding triangular numbers divisible by 5)

Use the above solutions for $(10x+6)^2=100k+10x+6$ and so on for other solutions.