2025 AMC 10A Problems/Problem 17: Difference between revisions

Revision as of 20:16, 7 November 2025

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. $\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45$ by the Euclidean Algorithm, so the answer is $\boxed{\text{(E) }4}.$

Note: If an integer $a$ is congruent to an integer $b$ modulo a positive integer $c$ , denoted by $a \equiv b \pmod{c}$ , this means that $c$ divides the difference $a-b$

~Tacos_are_yummy_1

minor edit by AD_12

Solution 2

We are given that $273436 \equiv 16 \pmod N$ and $272760 \equiv 15 \pmod N$ , from which we get $273420 \equiv 0 \pmod N$ and $272745 \equiv 0 \pmod N$

Substracting the two gives $273420 - 272745 = 675$ , which is also divisible by $N$ .

Since $N$ must be greater than $16$ , the only possible divisors of $675$ are $25, 27, 45, 75, 135, 225,$ and $675$ . Checking which ones also divide $273436$ and $273420$ eliminates $25$ and $27$ , but $273420 \equiv 0 \pmod {45}$ .

Since both $273420$ and $675$ are divisible by $45$ , so is $273420-675=272745$ . Thus, $N = 45$ works, and its tens digit is $\boxed{\text{(E) }4}$ .

~Continuous_Pi

Solution 3

We get that: $273436 \equiv {16}\pmod{N}$ and $272760 \equiv {15}\pmod{N}$ .

So we also have that: $273420 \equiv {0}\pmod{N}$ and $272745 \equiv {0}\pmod{N}.$

Notice that these are a multiple of $5$ . Now, we subtract these numbers to get $675$ . We see that $675 = 25 \cdot 27.$ There is a factor of $5$ and $9$ . $273420$ and $272745$ are multiples of $9$ as well, so our answer is just 45, or $\boxed{\text{(E) }4}$ .

~Aarav22

~reformatting by Alzwang

~minor editing by kfclover

Video Solution (In 2 Mins)

https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 46: / Line 46: @@
 ~minor editing by kfclover
-==Solution 4 (Extremely Roundabout)==
 == Video Solution (In 2 Mins) ==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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