Indian Statistical Institute B.Math & B.Stat : Number Theory

Find the digit at the unit place of $$\big(1!-2!+3!-\dots \dots +25!\big )^{\big(1!-2!+3!-\dots \dots +25!\big )}$$ First note that $k! \equiv 0 (mod\ 10)$ for all $k \geq 5 , k \in \mathbb{N}$ So, $5!-6!+7!-\dots \dots +25! \equiv 0 (mod\ 10)$ and $1!-2!+3!-4! = -19 \equiv 1 (mod\ 10)$ (Using the property of $congruences$). Using the above two congruences $\big(1!-2!+3!-\dots \dots +25!\big ) \equiv 1 (mod\ 10)$ So, $$\big(1!-2!+3!-\dots \dots +25!\big )^{\big(1!-2!+3!-\dots \dots +25!\big )} \equiv 1^{\big(1!-2!+3!-\dots \dots +25!\big )} \equiv 1 (mod 10)$$ giving $1$ as the last digit. Let $a \equiv a' (mod\ m)$ and $b \equiv b' (mod\ m)$, then important properties of $congruences$ include the following, where $\implies$ means "implies": 1. Reflexivity: $a\equiv a (mod- m)$. 2. Symmetry: $a\equiv b (mod\ m) \implies b\equiv a (mod\ m)$. 3. Transitivity: $a\equiv b (mod\ m)$ and $b \equiv c (mod\ m)\implies a\equiv c (mod\ m)$. 4. $a+b \equiv a'+b' (mod\ m)$ 5. $a-b\equiv a'-b' (mod\ m)$. 6. $ab\equiv a'b' (mod\ m)$. 7. $a\equiv b (mod\ m)\implies ka \equiv kb (mod\ m)$. 8. $a\equiv b (mod\ m)\implies a^n\equiv b^n (mod\ m)$. 9. $ak\equiv bk (mod\ m)\implies$ $a\equiv b \big(mod\ \frac{m}{(k,m)}\big),$ where $(k,m)$ is the greatest common divisor. 11. If $a \equiv b (mod\ m)$, then $P(a) \equiv P(b) (mod\ m)$, for $P(x)$ a polynomial with integer coefficients.