1. ## Congruence

Suppose $a,b,c,d$ are integers with $0 \le a \le b \le 99, 0 \le c \le d \le 99$. For any integer $i$, let $n_i = 101i+1002^i$. Show that if $n_a + n_b$ is congruent to $n_c+n_d \pmod {10100}$ then $a=c$ and $b=d$.

2. I will at least rewrite the problem for you. Consider it a favor (since the Ravens are in the playoffs!). Others may now actually look at your problem.

Suppose $\displaystyle a, b, c, d$ are [non-negative] integers satisfying the conditions
$\displaystyle 0 \leq a \leq b \leq 99$
and
$\displaystyle 0 \leq c \leq d \leq 99$.

For any integer $\displaystyle i \text{, let } n_i = 101i + 1002^i$.

Show that if
$\displaystyle n_a + n_b \equiv n_c + n_d \pmod{10100}$
then
$\displaystyle a=c \text{ and } b=d$.

3. I'll give you a start.

First, 10100 is huge. So let's break it down into some of its relatively prime factors. 101, 25, and 4 should do the trick. Note that 101*25*4 = 10100.

Thus,

$\displaystyle n_a + n_b \equiv n_c + n_d \pmod{10100} \implies$

$\displaystyle n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{101}$

$\displaystyle n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{25}$

$\displaystyle n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{4}$