2025 AMC 8 Problems/Problem 16: Difference between revisions
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==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? | 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? | ||
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==Solution== | ==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>. | 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 | ~Soupboy0 | ||
One efficient method is to quickly add <math>(1, 2, 3, 4, 5, 6, 7, 8, 9, | == 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 | ~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]] | |||
Latest revision as of 12:35, 4 June 2025
Problem
Five distinct integers from 
to 
are chosen, and five distinct integers from 
to 
are chosen. No two numbers differ by exactly . What is the sum of the ten chosen numbers?
Solution
Note that for no two numbers to differ by , every number chosen must have a different units digit. To make computations easier, we can choose 
from the first group and 
from the second group. Then the sum evaluates to 
.
~Soupboy0
Another Way To Compute
For 
, we can add the first term and the last term, which is 
. If we add the second term and the second-to-last term it is also . There are 
pairs that sum to , so the answer is 
which equals 
.
- leafy
Similar solution
One efficient method is to quickly add 
, which is 
. Then because you took 
in total away from , you add . 
.
~Bepin999
Solution 4 (effecient)
To solve this problem, I started with the easiest/smallest case possible. In my opinion, that was
.
I then solved this equation quickly using Little Gauss's method, rearranging that into
.
I simplified this into
.
Solving this simple equation gives us the answer, which is .
~Kapurnicus
~Minor edit by NYCnerd
Video Solution(Quick, fast, easy!)
~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
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
