#nt
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$N = a_n a_{n-1} a_{n-2} \dots a_1 a_0 \quad \text{where} \quad N = 10^{n}a_n + 10^{n-1}a_{n-1} + 10^{n-2}a_{n-2} + \dots + 10^{1}a_1 + 10^{0}a_0$
Cσrd
10 = 2mod4 and 1 = 1mod4 so 10a0 + a1 = (2a1 +a0) mod 4
10^n = 0mod4 where n>= 2 so it entirely depends on the last 2 digits being divisible by 4
thank youuu