Suppose are integers with . For any integer , let . Show that if is congruent to then and .
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 $.
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
$
$\displaystyle
n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{101}$
$\displaystyle
n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{25}$
$\displaystyle
n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{4}$