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

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==Problem==
==Problem==
Line 5: Line 4:


<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. Hence, we can just find the greatest common divisor of the two numbers. <imath>\gcd(273420,272745)=\gcd(675,272745)</imath> Which simplifies to <imath>\gcd(675,45)=45,</imath> by the Euclidean Algorithm so the answer is <imath>\boxed{\text{(E) }4}.</imath>
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>
 
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


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


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


First we substract (273436 − 272760) = 676 ≡ 1 (mod N)
Substracting the two gives <imath>273420 - 272745 = 675</imath>, which is also divisible by <imath>N</imath>.


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 <imath>\boxed{\text{(E) }4}.</imath>.
Since <imath>N</imath> must be greater than <imath>16</imath>, the only possible divisors of <imath>675</imath> are <imath>25, 27, 45, 75, 135, 225,</imath> and <imath>675</imath>. Checking which ones also divide <imath>273436</imath> and <imath>273420</imath> eliminates <imath>25</imath> and <imath>27</imath>, but <imath>273420 \equiv 0 \pmod {45}</imath>.
 
Since both <imath>273420</imath> and <imath>675</imath> are divisible by <imath>45</imath>, so is <imath>273420-675=272745</imath>. Thus, <imath>N = 45</imath> works, and its tens digit is <imath>\boxed{\text{(E) }4}</imath>.


~Continuous_Pi
~Continuous_Pi


==Alternate Solution==
~edited by [[User:Zhixing|Zhixing]]
We are given \(273436 \equiv 16 \pmod{N}\) and \(272760 \equiv 15 \pmod{N}.\) 
 
Subtracting the second equation from the first yields \(273436 - 272760 \equiv 16 - 15 \pmod{N}.\) 
==Solution 3==
so now, \(676 \equiv 1 \pmod{N}.\)  
 
Hence, \(N \mid (676 - 1) = 675.\) 
We get that:
The prime factorization of 675 is \(675 = 5^2 \times 3^3.\)  
<imath>273436 \equiv {16}\pmod{N}</imath> and
Therefore, \(N\) must be a divisor of 675.   
<imath>272760 \equiv {15}\pmod{N}</imath>.
Since the remainders are less than \(N,\) we have \(N > 16.\)  
 
The largest divisor of 675 greater than 16 is \(45.\)  
So we also have that:
The tens digit of 45 is 4. 
<imath>273420 \equiv {0}\pmod{N}</imath> and
Thus, the answer is \(\boxed{\text{(E) }4}.\)
<imath>272745 \equiv {0}\pmod{N}.</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>.
<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
 
~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) ==
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


~WildSealVM/Vincent M.
==See Also==
{{AMC10 box|year=2025|ab=A|num-b=16|num-a=18}}
* [[AMC 10]]
* [[AMC 10 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}

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

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