2025 AMC 10A Problems/Problem 17

(Problem goes here)

Problem

Let $N$ be the unique positive integer such that dividing $273436$ by $N$ leaves a remainder of $16$ and dividing $272760$ by $N$ leaves a remainder of $15$ . What is the tens digit of $N$ ?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution 1

The problem statement implies $N|273420$ and $N|272745.$ We want to find $N > 16$ that satisfies both of these conditions. Hence, we can just find the greatest common divisor of the two numbers. $\gcd(273420,272745)=\gcd(675,272745)$ Which simplifies to $\gcd(675,45)=45,$ by the Euclidean Algorithm so the answer is $\boxed{\text{(E) }4}.$

~Tacos_are_yummy_1

Solution 2

We have: 273436 ≡ 16 (mod N) 272760 ≡ 15 (mod N)

First we substract (273436 − 272760) = 676 ≡ 1 (mod N)

So N divides 675. Since N is greater than 16, possible divisors are all greater than 16 and are 25, 27, 45, 75, 135, 225, 675. We then check which ones work. If 273436 ≡ 16 (mod N), then 273420 must be divisible by N. 273420 ÷ 45 = 6076, so N = 45 works. So N = 45, and the tens digit is $\boxed{\text{(E) }4}.$ .

~Continuous_Pi

Alternate Solution

We are given $273436 \equiv 16 \pmod{N}$ and $272760 \equiv 15 \pmod{N}.$ Subtracting the second equation from the first yields $273436 - 272760 \equiv 16 - 15 \pmod{N}.$ so now, $676 \equiv 1 \pmod{N}.$ Hence, $N \mid (676 - 1) = 675.$ The prime factorization of 675 is $675 = 5^2 \times 3^3.$ Therefore, $N$ must be a divisor of 675. Since the remainders are less than $N,$ we have $N > 16.$ The largest divisor of 675 greater than 16 is $45.$ The tens digit of 45 is 4. Thus, the answer is $\boxed{\text{(E) }4}.$

~WildSealVM/Vincent M.

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2025 AMC 10A Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

Alternate Solution