Difference between revisions of "2017 AMC 8 Problems/Problem 12"
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| − | ==Problem | + | ==Problem== |
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers? | The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers? | ||
<math>\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124</math> | <math>\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | The LCM(4,5,6) is 60. Since <math>60+1=61</math>, and that is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math> | + | Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder.The <math>\operatorname{LCM}(4,5,6)</math> is <math>60</math>. Since <math>60+1=61</math>, that is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math> |
| + | |||
| + | ==Solution 2== | ||
| + | Call the number we want to find <math>n</math>. We can say that | ||
| + | <cmath>n \equiv 1 \mod 4</cmath> | ||
| + | <cmath>n \equiv 1 \mod 5</cmath> | ||
| + | <cmath>n \equiv 1 \mod 6.</cmath> | ||
| + | We can also say that <math>n-1</math> is divisible by <math>4,5</math>, and <math>6.</math> | ||
| + | Therefore, <math>n-1=lcm(4,5,6)=60</math>, so <math>n=60+1=61</math> which is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math> | ||
| + | |||
| + | ~PEKKA | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/UIrHmHT_jl4 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/SwgXyYFIuxM | ||
| + | |||
| + | https://youtu.be/kNIx_iDhVD4 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | https://www.youtube.com/watch?v=nOSqQjEv0U0 ~David | ||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 19:49, 7 June 2025
Contents
Problem
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?
Solution 1
Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder.The 
is 
. Since 
, that is in the range of 
Solution 2
Call the number we want to find 
. We can say that
![\[n \equiv 1 \mod 4\]](http://latex.artofproblemsolving.com/2/5/3/253b4a316681500502a2cfca6ee6ba8005a8dc02.png)
![\[n \equiv 1 \mod 5\]](http://latex.artofproblemsolving.com/1/b/d/1bd82797818d62bf375e80ff0a90a2c61d8a51f8.png)
![\[n \equiv 1 \mod 6.\]](http://latex.artofproblemsolving.com/2/d/8/2d89f62bc17793757f37ac1a06d81cafce7ea80e.png)
We can also say that 
is divisible by 
, and 
Therefore, 
, so 
which is in the range of
~PEKKA
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
https://www.youtube.com/watch?v=nOSqQjEv0U0 ~David
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
