Given below are two statements Statement I: A number is divisible by 25, if…

Concept: For a divisor d that exactly divides some power of ten (10k), the divisibility of any whole number N by d depends only on the number formed by N's last k digits. Writing N = (leading digits) × 10k + (last k digits), the leading-digit term is always a multiple of d whenever d divides 10k exactly — so only the last-k-digit block decides divisibility.

Application:

1. Statement I (divisor 25): 102 = 100, and 100 = 25 × 4 is exactly divisible by 25. So for any number N, N = (leading digits) × 100 + (last two digits); the leading-digit term is always a multiple of 25, meaning N is divisible by 25 precisely when its last two digits form 00 or a multiple of 25 (25, 50, or 75) — confirming Statement I states a valid rule.

2. Statement II (divisor 16): 104 = 10000, and 10000 = 16 × 625 is exactly divisible by 16. So for any number N, N = (leading digits) × 10000 + (last four digits); the leading-digit term is always a multiple of 16, meaning N is divisible by 16 precisely when the number formed by its last four digits is divisible by 16 — confirming Statement II states a valid rule.

Cross-check:

• For Statement I, take N = 1350: its last two digits form 50, which is divisible by 25 (50 = 25 × 2); direct division confirms 1350 ÷ 25 = 54, matching the shortcut.

• For Statement II, take N = 123408: its last four digits form 3408, and 3408 ÷ 16 = 213 exactly; direct division confirms 123408 ÷ 16 = 7713, matching the shortcut again.

Result: Both Statement I and Statement II state correct, standard divisibility rules, each following directly from the fact that the relevant power of ten — 100 for 25, and 10000 for 16 — is itself exactly divisible by that number.