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

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<imath>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</imath>
<imath>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</imath>
==Video Solution==
https://youtu.be/CCYoHk2Af34


==Solution 1==
==Solution 1==
The problem statement implies <imath>N|273420</imath> and <imath>N|272745.</imath> We want to find <imath>N > 16</imath> that satisfies both of these conditions.  
The problem statement implies that <imath>N</imath> divides both <imath>273436-16=273420</imath> and <imath>272760-15=272745</imath>. We want to find <imath>N > 16</imath> that satisfies both of these conditions.  
Hence, we can just find the greatest common divisor of the two numbers. <imath>\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45</imath> by the Euclidean Algorithm, so the answer is <imath>\boxed{\text{(E) }4}.</imath>
Hence, we can just find the greatest common divisor of the two numbers. <imath>\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45</imath> by the Euclidean Algorithm, so the answer is <imath>\boxed{\text{(E) }4}.</imath>


Note: <imath></imath>\text{\fbox{ \parbox{0.95\textwidth}{ \textbf{Definition (Modular Congruence):} If an integer <imath>a</imath> is congruent to an integer <imath>b</imath> modulo a positive integer <imath>c</imath>, denoted by <imath>a \equiv b \pmod{c}</imath>, this means that <imath>c</imath> divides the difference <imath>a-b</imath>
Note: If an integer <imath>a</imath> is congruent to an integer <imath>b</imath> modulo a positive integer <imath>c</imath>, denoted by <imath>a \equiv b \pmod{c}</imath>, this means that <imath>c</imath> divides the difference <imath>a-b</imath>


~Tacos_are_yummy_1
~Tacos_are_yummy_1
~andliu766
minor edit by AD_12
minor edit by AD_12


==Solution 2==
==Solution 2==


We are given that <imath>273436 \equiv 16 \pmod N</imath> and <imath>272760 \equiv 15 \pmod N</imath>, from which we get <imath>273420 \equiv 0 \pmod N</imath> and <imath>272745 \equiv 0 \pmod N</imath>
We are given that <imath>273436 \equiv 16 \pmod N</imath> and <imath>272760 \equiv 15 \pmod N</imath>, from which we get <imath>273420 \equiv 0 \pmod N</imath> and <imath>272745 \equiv 0 \pmod N</imath>.


Substracting the two gives <imath>273420 - 272745 = 675</imath>, which is also divisible by <imath>N</imath>.
Substracting the two gives <imath>273420 - 272745 = 675</imath>, which is also divisible by <imath>N</imath>.
Line 25: Line 31:


~Continuous_Pi
~Continuous_Pi
~edited by [[User:Zhixing|Zhixing]]


==Solution 3==
==Solution 3==
Line 38: Line 46:
Notice that these are a multiple of <imath>5</imath>. Now, we subtract these numbers to get <imath>675</imath>.
Notice that these are a multiple of <imath>5</imath>. Now, we subtract these numbers to get <imath>675</imath>.
We see that <imath>675 = 25 \cdot 27.</imath> There is a factor of <imath>5</imath> and <imath>9</imath>.
We see that <imath>675 = 25 \cdot 27.</imath> There is a factor of <imath>5</imath> and <imath>9</imath>.
<imath>273420</imath> and <imath>272745</imath> are multiples of <imath>9</imath> as well, so our answer is just <imath>\boxed{\text{(E) }45}</imath>.
<imath>273420</imath> and <imath>272745</imath> are multiples of <imath>9</imath> as well, so our answer is just 45, or <imath>\boxed{\text{(E) }4}</imath>.


~Aarav22
~Aarav22


~reformatting by Alzwang
~reformatting by Alzwang
~minor editing by kfclover
==Solution 4==
We are given that <imath>273436 \equiv 16 \pmod{N}</imath> and <imath>272760 \equiv 15 \pmod{N}.</imath> 
Subtracting, we get <imath>273436 - 272760 = 676 \equiv 1 \pmod{N}.</imath> 
This means <imath>N</imath> divides <imath>675.</imath> 
Also, since <imath>273436 \equiv 16 \pmod{N},</imath> we have <imath>273420 \equiv 0 \pmod{N}.</imath> 
Thus <imath>N</imath> divides both <imath>273420</imath> and <imath>675.</imath> Using the Euclidean Algorithm: 
<imath>273420 - 405 \times 675 = 45,</imath> so <imath>\gcd(273420, 675) = 45.</imath> 
Hence <imath>N = 45,</imath> and the tens digit of <imath>N</imath> is <imath>\boxed{\text{(E) }4}.</imath>
~samma
==Chinese Video Solution==
https://www.bilibili.com/video/BV1nLkQBpEXR/
~metrixgo


== Video Solution (In 2 Mins) ==
== Video Solution (In 2 Mins) ==
https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy
https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==Video Solution==
==Video Solution==

Latest revision as of 13:02, 10 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 $N$?

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

Video Solution

https://youtu.be/CCYoHk2Af34

Solution 1

The problem statement implies that $N$ divides both $273436-16=273420$ and $272760-15=272745$. 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

~andliu766

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

~edited by Zhixing

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

Solution 4

We are given that $273436 \equiv 16 \pmod{N}$ and $272760 \equiv 15 \pmod{N}.$ Subtracting, we get $273436 - 272760 = 676 \equiv 1 \pmod{N}.$

This means $N$ divides $675.$ Also, since $273436 \equiv 16 \pmod{N},$ we have $273420 \equiv 0 \pmod{N}.$

Thus $N$ divides both $273420$ and $675.$ Using the Euclidean Algorithm: $273420 - 405 \times 675 = 45,$ so $\gcd(273420, 675) = 45.$

Hence $N = 45,$ and the tens digit of $N$ is $\boxed{\text{(E) }4}.$

~samma

Chinese Video Solution

https://www.bilibili.com/video/BV1nLkQBpEXR/

~metrixgo

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

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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