# Square-root of CAREER alphametic

Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.

$\sqrt{CAREER} \,\, = \,\, RUT$

Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)

Since squares can only be $0, 1, 4, 5, 6, 9 \mod 10$, $R$ is one of those numbers. Further, since a three-digit number has a six-digit square, we have $R \geq 3$. So $R$ is $4, 5, 6$ or $9$.

However, since $T \neq R$, $R$ can't be $5$. This leaves $4$, $6$ or $9$ for $R$.

Case 1: $R=4$. Then $C=1$. Also, since the square of $T$ ends in a 4, $T$ is $2$ or $8$.

Case 1.1. $R=4$, $C=1$, $T=2$: $$104004+10000A+110E=161604+8040U+100U^2$$ $$10000A+110E=57600+8040U+100U^2$$

Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 and 2 are used we get $U=3$. Then $10000A+110E=82620$, which gives no solution for $A$ and $E$.

Case 1.2. $R=4$, $C=1$, $T=8$: $$104004+10000A+110E=166464+8160U+100U^2$$ $$10000A+110E=62460+8160U+100U^2$$

Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 is used we get $U=2$ or $U=3$. Then $10000A+110E=95940$ or $10000A+110E=82870$, neither of which gives no solution for $A$ and $E$.

Case 2. $R=6$. Since the square of $T$ ends in a 6, $T$ is $4$ or $6$, but 6 is taken. So $R=6$ and $T=4$. We look to the fourth digit of $(6U4)^2$, which is $12U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor$, but also $R=6$.

$$2U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor \equiv 6 \mod 10$$

$U=1$ is possible. This gives $614^2=376996$, which is in fact an acceptable solution. $U=2$ isn't possible, since this gives $9 \not\equiv 6 \mod 10$. $U=3$ isn't possible, since this gives $12 \not\equiv 6 \mod 10$. 4 and 6 are taken. $U=7$ isn't possible, since this gives $24 \not\equiv 6 \mod 10$. $U=8$ isn't possible, since this gives $27 \not\equiv 6 \mod 10$. $U=9$ isn't possible, since this gives $31 \not\equiv 6 \mod 10$.

Case 3. $R=9$. Since the square of $T$ ends in a 9, $T$ is $3$ or $7$.

Case 3.1. $R=9$, $T=3$. We look to the fourth digit of $(9U3)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.

So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$

$$8U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$

$U=1$ isn't possible, since this gives $13 \not\equiv 9 \mod 10$. $U=2$ isn't possible, since this gives $21 \not\equiv 9 \mod 10$. $3$ is already taken. $U=4$ is possible. However $943^2=889249$, which doesn't match the given form. $U=5$ isn't possible, since this gives $48 \not\equiv 9 \mod 10$. $U=6$ isn't possible, since this gives $57 \not\equiv 9 \mod 10$. $U=7$ isn't possible, since this gives $66 \not\equiv 9 \mod 10$. $U=8$ isn't possible, since this gives $76 \not\equiv 9 \mod 10$. 9 is already taken.

Case 3.2. $R=9$, $T=7$. We look to the fourth digit of $(9U7)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.

So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$

$$8U+\left\lfloor \frac{12600+100U^2+140U}{1000} \right\rfloor \equiv 9 \mod 10$$

$$8U+\left\lfloor \frac{2600+100U^2+140U}{1000} \right\rfloor \equiv 8 \mod 10$$

$U=1$ isn't possible, since this gives $10 \not\equiv 8 \mod 10$. $U=2$ isn't possible, since this gives $19 \not\equiv 8 \mod 10$. $U=3$ isn't possible, since this gives $27 \not\equiv 8 \mod 10$. $U=4$ isn't possible, since this gives $36 \not\equiv 8 \mod 10$. $U=5$ isn't possible, since this gives $45 \not\equiv 8 \mod 10$. $U=6$ isn't possible, since this gives $54 \not\equiv 8 \mod 10$. $U=8$ isn't possible, since this gives $74 \not\equiv 8 \mod 10$. 7 and 9 are already taken.

Hence the only solution is $C=3$, $A=7$, $E=9$, $R=6$, $U=1$ and $T=4$.

First of all, $R\geqslant 3$ since CAREER has 6 digit numbers. In addition, we know that CAREER is simply square of RUT, so R cannot be 3 since the end of CAREER digit is R and it is $RUT^2$, so R can be 4,5,9,6 (last digits of any square of a number cannot be other value of 0,1,4,5,6,9).

Moreover the R cannot be 5 since the last digit will be 5 and that will make T value is 5 which has to be different than R. So R can be 4,6,9.

If $R=6$, T has to be 4.

So CAREER becomes CA6EE6 and RUT becomes 6U4. The first digit of CAREER can only be 3 or 4. Since we have chosen T as 4. C can only be 3. So that makes U value as 1 or 2 because if we take $U>2$ that makes C as 4. As a result, so only 614 and 624 are possible option for $R=6$.

If $R=9$, T has to be 3 or 7 so;

If R=9 and T=3; our CAREER becomes CA9EE9 and RUT becomes 9U3. C can only be 8 since our R value has to be different than C. That narrows down the value of U as 1,2,3,4. and since T=3, U value can only be 1,2,4.

If R=9 and T=8; our CAREER becomes CA9EE9 and RUT becomes 9U7. C can only be 8 again. That narrows down the value of U as 1,2,3,4. U value can only be 1,2,3,4.

If $R=4$; T has to be 2 or 8.

If R=4 and T=2; our CAREER becomes CA4EE4 and RUT becomes 4U2. C can only be 1 or 2 and since T =2 C can only be 1. That narrows U values down to 1,2,3,4. And C is 1 and T is 2, R is 4, U can only be 3.

Since it is narrowed down quite a bit, we can try and find;

CAREER = 376996 and RUT=614