Let $p(x)= a_{n}x^n +a_{n-1}x^{n-1} \dots + a_{1}x+a_{0} $ be a polynomial with integer coefficients. So we have
$$ 2 | P(5) = a_{n}5^n +a_{n-1}5^{n-1} \dots + a_{1}5+a_{0}, \ \text{and} \tag{A} $$
$$ 5| P(2)= a_{n}2^n +a_{n-1}2^{n-1} \dots + a_{1}2+a_{0}, \ \tag{B} $$
Also, we have
$$ P(7)= a_{n}7^n +a_{n-1}7^{n-1} \dots + a_{1}7+a_{0} $$
Since $7 \equiv 2 (\mod{5}), $ therefore we have
$$ P(7) \equiv a_{n}2^n +a_{n-1}2^{n-1} \dots + a_{1}2+a_{0} (\mod{5}) $$
So using $(B),$ we have $5$ divides $P(7).$ Similarly, $7\equiv 5 (\mod{2}) $
$$ P(7) \equiv a_{n}5^n +a_{n-1}5^{n-1} \dots + a_{1}5+a_{0} (\mod{2}) $$
Using $(A)$ we have $2$ divides $P(7).$ Therefore, $10$ divides $P(7).$
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