The Power of All Powers

Since $7^{abcd} \equiv 7$ modulo 11, it follows that $abcd \equiv 1$ modulo 10. Thus $a,b,c,d$ are all odd, and not multiples of $5$ . Since $c$ is odd, $d \equiv 2^c \equiv 2$ modulo 3. Since $d$ is odd, this means that $d \equiv 5$ modulo 6.

Since $b$ is odd and $3^2 \equiv -1$ modulo 5, we see that $c \equiv 3^b \equiv \pm 3$ modulo 5. Since $c$ is odd, this implies that $c \equiv 3,7$ modulo 10. If $c \equiv 3$ modulo 10, then $3^b \equiv c \equiv 3$ modulo 5, and hence $b \equiv 1$ modulo 4. If $c \equiv 7$ modulo 10, then $3^b \equiv c \equiv 2$ modulo 5, and hence $b \equiv 3$ modulo 4.

Thus $d \equiv 5$ modulo 6 and either $c \equiv 3$ modulo 10 and $b \equiv 1$ modulo 4, or else $c \equiv 7$ modulo 10 and $b \equiv 3$ modulo 4.

Trying $d=11$ , $c=7$ , we quickly obtain $b=3$ and $a = 11$ as one particular solution. Let us try to find a solution with a product smaller than $11 \times 3 \times 7 \times 11 = 2541$ .

If we look for solutions with $d=11$ , then $a,b,c$ must be odd numbers greater than $1$ and not equal to $5$ whose product is less than $231$ . Since $7^3 = 343$ , at least one of these numbers must be equal to $3$ . The only possible choice of numbers containing just a single $3$ is $3,7,7$ . This option is no good, since their product is not congruent to $1$ modulo $10$ (and $abc \equiv 11abc \equiv abcd \equiv 1$ modulo 10). Thus $a,b,c$ must consist of two $3$ s and an odd number congruent to $9$ modulo 10. The only possibilities are $3,3,9$ and $3,3,19$ . Since $5^3 \equiv 6$ modulo 7 and none of $3,9,19$ are congruent to $6$ modulo 7, $a$ must be either $9$ or $19$ , which would require $b=c=3$ . This is impossible, since $3^3 \not\equiv 3$ modulo 5. We can obtain no product $abcd$ less than $2541$ with $d=11$ .

If we look for solutions with $d > 11$ , then $d \ge 17$ and hence $abc < 149$ . Thus $a,b,c$ must include at least one $3$ (as above), and the only combination with just one $3$ is $3,7,7$ . Since $5^3 \equiv 6 \not\equiv 7$ modulo 7 and $5^7 \equiv 5 \not\equiv 3,7$ modulo 7, this combination is not possible. Thus the collection $a,b,c$ must contain at least two $3$ s, and so must be $3,3,k$ where $k$ is one of the odd numbers $3,7,9,11,13$ . Since $3^3 \not\equiv 3$ modulo 5, we cannot have $b=c=3$ . Thus we must have $a=3$ , and hence $b \equiv 5^a \equiv 6$ modulo 7, so we must have $b=13$ . With $a=3$ , $b=13$ and $c=3$ , we must have $d \le 21$ and hence $d = 17$ . But $3\times13\times3\times17 \not\equiv1$ modulo $10$ .

There are no solutions with a smaller product than $2541$ .

According to Fermat's little theorem, if n is not divisible by prime p, then $n^p$ ≡ $n$ (mod $p$ )

or $n^{p-1}$ ≡ $1$ (mod $p$ ).

From the last constraint, $7^{abcd}$ ≡ 7 (mod 11); $7^{abcd-1}$ ≡ 1 (mod 11).

That means $abcd-1$ is a multiple of p-1 = 11-1 =10, for $n^{10}$ ≡ $1$ (mod $11$ ). In other words, the product $abcd$ must end with a digit 1 as 1 is a remainder of the divisor 10. Furthermore, we can conclude that $a$ , $b$ , $c$ , $d$ must be odd integers that do not end with 5 as 5 only results in a product ending with 5.

Now let's consider $2^c$ ≡ $d$ (mod 3). Clearly, $d$ ≡ 1 or 2 (mod 3) because the power of 2 can never be divided by 3. However, if $d$ ≡ 1 (mod 3), then

$2^c$ ≡ 1 (mod 3), which again correlates with the formula $n^{p-1}$ ≡ $1$ (mod $p$ ).

Hence, c is a multiple of p-1 = 3-1 = 2, which is contradicted because $c$ must be odd.

Thus, $\boxed{d\ ≡ 2 (mod\ 3)}$ . Then $2^c$ ≡ 2 (mod 3); $2^{c-1}$ ≡ 1 (mod 3); c-1 ≡ 0 (mod 2); $\boxed{c\ ≡ 1 (mod\ 2)}$ .

Then $c$ can have a remainder of 1, 2, 3, or 4 when divided by 5. However, if c ≡ 1 (mod 5), then

$3^b$ ≡ 1 (mod 5); b ≡ 0 (mod 4), which is even and so contradicted.

Also, if c ≡ 4 (mod 5), then $3^b$ ≡ 4 (mod 5); $3^6$ ≡ 4 (mod 5); $3^{b-6}$ ≡ 1 (mod 5); b - 6 = 0 (mod 4), which is even and contradicted.

Now if c = 2 (mod 5), then $3^b$ ≡ 2 (mod 5); $3^3$ ≡ 2 (mod 5); $3^{b-3}$ ≡ 1 (mod 5); b-3 ≡ 0 (mod 4) or $b$ ≡ 3 (mod 4).

And if c = 3 (mod 5), then $3^b$ ≡ 3 (mod 5); $3^{b-1}$ ≡ 1 (mod 5); b-1 ≡ 0 (mod 4) or $b$ ≡ 1 (mod 4).

Thus, $\boxed{c\ ≡ 2 (mod\ 5)}$ or $\boxed{c\ ≡ 3 (mod\ 5)}$ .

So now we have 2 possible remainders for $c$ . We will move on to possibility of $b$ .

Similarly, $b$ can have a remainder of 1, 2, 3, 4, 5, or 6 when divided by 7. However, if b ≡ 1 (mod 7), then

$5^a$ ≡ 1 (mod 7); a ≡ 0 (mod 6), which is even and contradicted.

If b ≡ 2 (mod 7), then $5^a$ ≡ 2 (mod 5); $5^4$ ≡ 2 (mod 7); $5^{a-4}$ ≡ 1 (mod 7); a-4 ≡ 0 (mod 6), which is even and contradicted.

If b ≡ 4 (mod 7), then $5^a$ ≡ 4 (mod 7); $5^2$ ≡ 4 (mod 7); $5^{a-2}$ ≡ 1 (mod 7); a-2 ≡ 0 (mod 6), which is even and contradicted.

If b ≡ 3 (mod 7), then $5^a$ ≡ 3 (mod 7); $5^5$ ≡ 3 (mod 7); $5^{a-5}$ ≡ 1 (mod 7); a-5 ≡ 0 (mod 6); a ≡ 5 (mod 6).

If b ≡ 5 (mod 7), then $5^a$ ≡ 5 (mod 7); $5^{a-1}$ ≡ 1 (mod 7); a-1 ≡ 0 (mod 6); a ≡ 1 (mod 6).

If b ≡ 6 (mod 7), then $5^a$ ≡ 6 (mod 7); $5^3$ ≡ 6 (mod 7); $5^{a-3}$ ≡ 1 (mod 7); a-3 ≡ 0 (mod 6); a ≡ 3 (mod 6).

Thus, $\boxed{b\ ≡ 3 (mod\ 7)}$ & $\boxed{a\ ≡ 5 (mod\ 6)}$ ; or $\boxed{b\ ≡ 5 (mod\ 7)}$ & $\boxed{a\ ≡ 1 (mod\ 6)}$ ; or $\boxed{b\ ≡ 6 (mod\ 7)}$ & $\boxed{a\ ≡ 3 (mod\ 6)}$ .

Finally, we have 3 possible remainders of modulus 7 of b, which will lead to a specific modulus of $a$ , and when combined with c possibility, we will get 6 possible cases:

Case 1 : $d$ ≡ 2 (mod 3); $c$ ≡ 2 (mod 5) ≡ 1 (mod 2); $b$ ≡ 3 (mod 4) ≡ 3 (mod 7); $a$ ≡ 5 (mod 6)

Obviously, the least value of possible $d$ = 11 as 5 can't be used. Then $c$ = 7. $b$ = 3, and $a$ = 11. (Note that the product must end with digit 1) Thus, $abcd$ in this case equals to 2541.

Case 2 : $d$ ≡ 2 (mod 3); $c$ ≡ 2 (mod 5) ≡ 1 (mod 2); $b$ ≡ 3 (mod 4) ≡ 5 (mod 7); $a$ ≡ 1 (mod 6)

Again, the least value of possible $d$ = 11 as 5 can't be used. Then $c$ = 7. $b$ = 19, and $a$ = 7. The least product in this case = 10241.

Case 3 : $d$ ≡ 2 (mod 3); $c$ ≡ 2 (mod 5) ≡ 1 (mod 2); $b$ ≡ 3 (mod 4) ≡ 6 (mod 7); $a$ ≡ 3 (mod 6)

Again, the least value of possible $d$ = 11 as 5 can't be used. Then $c$ = 7. $b$ = 27, and $a$ = 9. The least product in this case = 18711.

Case 4 : $d$ ≡ 2 (mod 3); $c$ ≡ 3 (mod 5) ≡ 1 (mod 2); $b$ ≡ 1 (mod 4) ≡ 3 (mod 7); $a$ ≡ 5 (mod 6)

Again, the least value of possible $d$ = 11 as 5 can't be used. Then $c$ = 3. $b$ = 17, and $a$ = 11. The least product in this case = 6171.

Case 5 : $d$ ≡ 2 (mod 3); $c$ ≡ 3 (mod 5) ≡ 1 (mod 2); $b$ ≡ 1 (mod 4) ≡ 5 (mod 7); $a$ ≡ 1 (mod 6)

Again, the least value of possible $d$ = 17 as 5 can't be used. Then $c$ = 3. $b$ = 33, and $a$ = 7. The least product in this case = 11781.

Case 6 : $d$ ≡ 2 (mod 3); $c$ ≡ 3 (mod 5) ≡ 1 (mod 2); $b$ ≡ 1 (mod 4) ≡ 6 (mod 7); $a$ ≡ 3 (mod 6)

Again, the least value of possible $d$ = 11 as 5 can't be used. Then $c$ = 3. $b$ = 13, and $a$ = 9. The least product in this case = 3861.

As a result, case 1 results in the least product with values of $\boxed{d\ = 11}$ ; $\boxed{c\ = 7}$ ; $\boxed{b\ = 3}$ ; $\boxed{a\ = 11}$ ; and product of $\boxed{2541}$ .