Difference between revisions of "2017 AMC 8 Problems/Problem 12"
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<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== |
| − | 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>, 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== | ==Video Solution== | ||
https://youtu.be/SwgXyYFIuxM | https://youtu.be/SwgXyYFIuxM | ||
| + | |||
| + | https://youtu.be/kNIx_iDhVD4 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | https://www.youtube.com/watch?v=nOSqQjEv0U0 ~David | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:17, 15 April 2023
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
