Difference between revisions of "2025 AMC 8 Problems/Problem 16"

Latest revision as of 13:35, 4 June 2025

Problem

Five distinct integers from $1$ to $10$ are chosen, and five distinct integers from $11$ to $20$ are chosen. No two numbers differ by exactly $10$ . What is the sum of the ten chosen numbers?

$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 105 \qquad \textbf{(D)}\ 110 \qquad \textbf{(E)}\ 115$

Solution

Note that for no two numbers to differ by $10$ , every number chosen must have a different units digit. To make computations easier, we can choose $(1, 2, 3, 4, 5)$ from the first group and $(16, 17, 18, 19, 20)$ from the second group. Then the sum evaluates to $1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C)\ 105}}$ .

~Soupboy0

Another Way To Compute

For $1+2+3+4+5+16+17+18+19+20$ , we can add the first term and the last term, which is $21$ . If we add the second term and the second-to-last term it is also $21$ . There are $5$ pairs that sum to $21$ , so the answer is $21 \times 5$ which equals $\boxed{\text{(C)\ 105}}$ .

- leafy

Solution 4 (effecient)

To solve this problem, I started with the easiest/smallest case possible. In my opinion, that was

$1+2+3+4+5+16+17+18+19+20$ .

I then solved this equation quickly using Little Gauss's method, rearranging that into

$1+19+2+18+3+17+4+16+5+20$ .

I simplified this into

$20+20+20+20+25$ .

Solving this simple equation gives us the answer, which is $\boxed{\text{(C)\ 105}}$ .

~Kapurnicus

~Minor edit by NYCnerd

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

Video Solution by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

Video Solution 1 by SpreadTheMathLove

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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=DtG8sG4CK4RrUlVz&t=1815 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

@@ Line 1: / Line 1: @@
+==Problem==
 Five distinct integers from <math>1</math> to <math>10</math> are chosen, and five distinct integers from <math>11</math> to <math>20</math> are chosen. No two numbers differ by exactly <math>10</math>. What is the sum of the ten chosen numbers?
-<math>\hspace*{5mm}\text{(A) } 95 \quad \text{(B) } 100 \quad \text{(C) } 105 \quad \text{(D) } 110 \quad \text{(E) } 115</math>
+<math>\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 105 \qquad \textbf{(D)}\ 110 \qquad \textbf{(E)}\ 115</math>
+==Solution==
+Note that '''for no two numbers to differ by <math>10</math>, every number chosen must have a different units digit'''. To make computations easier, we can choose <math>(1, 2, 3, 4, 5)</math> from the first group and <math>(16, 17, 18, 19, 20)</math> from the second group. Then the sum evaluates to <math>1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C)\ 105}}</math>.
+~Soupboy0
+== Another Way To Compute ==
+For <math>1+2+3+4+5+16+17+18+19+20</math>, we can add the first term and the last term, which is <math>21</math>. If we add the second term and the second-to-last term it is also <math>21</math>. There are <math>5</math> pairs that sum to <math>21</math>, so the answer is <math>21 \times 5</math> which equals <math>\boxed{\text{(C)\ 105}}</math>.
+- leafy
+==Similar solution==
+One efficient method is to quickly add <math>(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)</math>, which is <math>55</math>. Then because you took <math>50</math> in total away from <math>(16, 17, 18, 19, 20)</math>, you add <math>50</math>. <math>55+50= \boxed{\text{(C)\ 105}}</math>.
+~Bepin999
+==Solution 4 (effecient)==
+To solve this problem, I started with the easiest/smallest case possible. In my opinion, that was
+<math>1+2+3+4+5+16+17+18+19+20</math>.
+I then solved this equation quickly using Little Gauss's method, rearranging that into
+<math>1+19+2+18+3+17+4+16+5+20</math>.
+I simplified this into
+<math>20+20+20+20+25</math>.
+Solving this simple equation gives us the answer, which is <math>\boxed{\text{(C)\ 105}}</math>.
+~Kapurnicus
+~Minor edit by NYCnerd
+==Video Solution(Quick, fast, easy!)==
+https://youtu.be/fdG7EDW_7xk
+~MC
+==Video Solution by Pi Academy==
+https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
+==Video Solution 1 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=jTTcscvcQmI
+==Video Solution (A Clever Explanation You’ll Get Instantly)==
+https://youtu.be/VP7g-s8akMY?si=DtG8sG4CK4RrUlVz&t=1815
+~hsnacademy
+==Video Solution by Thinking Feet==
+https://youtu.be/PKMpTS6b988
+==See Also==
+{{AMC8 box|year=2025|num-b=15|num-a=17}}
+{{MAA Notice}}
+[[Category:Introductory Number Theory Problems]]

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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