Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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== Solution 1 == | == Solution 1 == | ||
| − | The sum of all five numbers is <math>85</math>. Since <math>85</math> is <math>1</math> more than a multiple of <math>4</math>, the number subtracted must be <math>1</math> more than a multiple of <math>4</math>. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>. | + | The sum of all five numbers is <math>85</math>. Since <math>85</math> is <math>1</math> more than a multiple of <math>4</math>, the number being subtracted must be <math>1</math> more than a multiple of <math>4</math>. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>. |
~Gavin_Deng | ~Gavin_Deng | ||
| Line 14: | Line 14: | ||
~cxsmi | ~cxsmi | ||
| − | == Solution | + | == Solution 3 == |
| − | We try out every number using [[ | + | We try out every number using [[brute force]] and get <math>\boxed{\textbf{(C)}~17}</math>. |
| − | Note that this is not | + | Note that this is not practical and it is very time-consuming. |
| − | == Video Solution 1 by SpreadTheMathLove == | + | == Solution 4 == |
| + | |||
| + | Since the sum of <math>15 + \dots + 19</math> is odd, we immediately exclude B and D. We further note that if A is true, then E is true. Hence, the answer is C. | ||
| + | |||
| + | == Solution 5== | ||
| + | |||
| + | We can use the fact that <math>(a + b) \mod n = [(a \mod n) + (b \mod n)] \mod n</math>. Notice that <math>15, 16, 17, 18, 19</math> dividing by 4 have remainders <math>3, 0, 1, 2, 3</math>. Their sum is 9. It is easy to see that <math>9-1=8</math> is divisible by 4 and so C (17) is the correct answer. | ||
| + | |||
| + | == Video Solution 1 by Cool Math Problems == | ||
| + | |||
| + | https://youtu.be/BRnILzqVwHk?si=1KwyuFBUDqdMtC6t&t=2 | ||
| + | |||
| + | == Video Solution 2 by SpreadTheMathLove == | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | ==Video Solution | + | == Video Solution 3 == |
| − | + | ||
| − | ~hsnacademy | + | [//youtu.be/VP7g-s8akMY?si=eMjbNcrUSiSp30ND&t=324 ~hsnacademy] |
| − | == Video Solution | + | == Video Solution 4 by Thinking Feet == |
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==Video Solution by Daily Dose of Math== | + | == Video Solution 5 by Daily Dose of Math == |
| − | https://youtu.be/ | + | [//youtu.be/nkpdskFVgdM ~Thesmartgreekmathdude] |
| + | ==Video Solution(Quick, fast, easy!)== | ||
| + | https://youtu.be/fdG7EDW_7xk | ||
| − | ~ | + | ~MC |
== See Also == | == See Also == | ||
Latest revision as of 09:25, 16 July 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Video Solution 1 by Cool Math Problems
- 8 Video Solution 2 by SpreadTheMathLove
- 9 Video Solution 3
- 10 Video Solution 4 by Thinking Feet
- 11 Video Solution 5 by Daily Dose of Math
- 12 Video Solution(Quick, fast, easy!)
- 13 See Also
Problem
Sekou writes the numbers 
After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 
Which number did he erase?
Solution 1
The sum of all five numbers is 
. Since is 
more than a multiple of 
, the number being subtracted must be more than a multiple of . Thus, the answer is 
.
~Gavin_Deng
Solution 2
The sum of the residues of these numbers modulo is 
. Hence, the number being subtracted must be congruent to modulo . The only such answer is .
~cxsmi
Solution 3
We try out every number using brute force and get .
Note that this is not practical and it is very time-consuming.
Solution 4
Since the sum of 
is odd, we immediately exclude B and D. We further note that if A is true, then E is true. Hence, the answer is C.
Solution 5
We can use the fact that ![$(a + b) \mod n = [(a \mod n) + (b \mod n)] \mod n$](http://latex.artofproblemsolving.com/9/9/1/9912596152140f8dc08c47fe3f58d8cce38dcb10.png)
. Notice that 
dividing by 4 have remainders 
. Their sum is 9. It is easy to see that 
is divisible by 4 and so C (17) is the correct answer.
Video Solution 1 by Cool Math Problems
https://youtu.be/BRnILzqVwHk?si=1KwyuFBUDqdMtC6t&t=2
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 3
Video Solution 4 by Thinking Feet
Video Solution 5 by Daily Dose of Math
Video Solution(Quick, fast, easy!)
~MC
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
