2025 AMC 10A Problem 17

Remove the remainders first. The two conditions say exactly that
$$N \mid 273436 - 16 = 273420 \qquad\text{and}\qquad N \mid 272760 - 15 = 272745.$$
So $N$ divides any combination of these, in particular their difference:
$$273420 - 272745 = 675.$$

Now run the Euclidean algorithm on the pair:
$$\gcd(273420,\,272745) = \gcd(272745,\,675).$$
Since $675 \cdot 404 = 272700$, the next remainder is $272745 - 272700 = 45$, and $675 = 45 \cdot 15$, so
$$\gcd(273420,\,272745) = 45.$$

Hence $N$ is a divisor of $45$: one of $1, 3, 5, 9, 15, 45$. A remainder of $16$ requires $N > 16$, which leaves only
$$N = 45.$$

Check: $45 \cdot 6076 = 273420$, so $273436 = 45 \cdot 6076 + 16$; and $45 \cdot 6061 = 272745$, so $272760 = 45 \cdot 6061 + 15$. Both conditions hold, and the tens digit of $45$ is $4$.

The answer is $\boxed{\textbf{(E)}\ 4}$.

"Leaves remainder $r$" is a statement about divisibility of $a - r$, never about $a$ itself, and subtracting the remainder immediately is what makes the gcd usable. Once $N$ is known to divide a specific small number, the problem becomes a short list of divisors, and the inequality $N > \max(15, 16)$ that comes free with the word "remainder" picks out the unique one. The uniqueness promised by the problem is thus not an extra assumption but a consequence: $45$ is the only divisor of $\gcd$ that exceeds $16$.

Subtract the two given numbers but correct for the remainders: $273436 - 272760 = 676$, while the remainders differ by $16 - 15 = 1$, so $N \mid 676 - 1 = 675 = 3^3 \cdot 5^2$. Combine with $N \mid 273420 = 2^2 \cdot 3^2 \cdot 5 \cdot 7^2 \cdot 31$ to get $N \mid 3^2 \cdot 5 = 45$, and finish with $N > 16$ as before.

• Subtracting the two given numbers to get 676 and factoring that; because the remainders differ by 1 the correct difference is 675, and 676 wrongly suggests N = 26 with tens digit 2, choice (C).
• Forgetting that the divisor must exceed every remainder and accepting $N = 9$ or $N = 15$, giving tens digit $0$ or $1$.
• Reporting the units digit, or reading the tens digit of $45$ as $5$.