Find the number of all the 2-digit numbers satisfying the following congruences x = 3(mod7), x = 2(mod5).
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Use the Chinese remainder theorem.

Start with x = 3×5 + 2×7 = 15 + 14 = 29. Now,

• 29 ≡ 15 ≡ 1 (mod7)

• 29 ≡ 14 ≡ 4 (mod5)

Adjust for this by multiplying the first term in x by 3, and the second term by 3 (because 4×3 ≡ 12 ≡ 2 (mod5)).

So now x = 3×5×3 + 2×7×3 = 45 + 42 = 87, and

• 87 ≡ 45 ≡ 3 (mod7)

• 87 ≡ 42 ≡ 2 (mod5)

The CRT then says that x ≡ 87 (mod(7×5)) ≡ 87 (mod35), which is to say any number x = 87 + 35n satisfies both congruences (where n is any integer).

So there are 3 possible 2-digit numbers that work: {17, 52, 87}.

To confirm:

• 17 ≡ 15 + 2 ≡ 2 (mod5) and 17 ≡ 14 + 3 ≡ 3 (mod7)

• 52 ≡ 50 + 2 ≡ 2 (mod5) and 52 ≡ 49 + 3 ≡ 3 (mod7)

• 87 ≡ 85 + 2 ≡ 2 (mod5) and 87 ≡ 84 + 3 ≡ 3 (mod7)

0 is its x and y coordinate