Difference between revisions of "2015 AIME II Problems/Problem 1"

(Created page with "==Problem== Let <math>N</math> be the least positive integer that is both <math>22</math> percent less than one integer and <math>16</math> percent greater than another integ...")
 
Line 3: Line 3:
 
Let <math>N</math> be the least positive integer that is both <math>22</math> percent less than one integer and <math>16</math> percent greater than another integer. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Let <math>N</math> be the least positive integer that is both <math>22</math> percent less than one integer and <math>16</math> percent greater than another integer. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
==Solution==
+
==Solution 1==
  
 
If <math>N</math> is <math>22</math> percent less than one integer <math>k</math>, then <math>N=\frac{78}{100}k=\frac{39}{50}k</math>. In addition, <math>N</math> is <math>16</math> percent greater than another integer <math>m</math>, so <math>N=\frac{116}{100}m=\frac{29}{25}m</math>. Therefore, <math>k</math> is divisible by 50 and <math>m</math> is divisible by 25. Setting these two equal, we have <math>\frac{39}{50}k=\frac{29}{25}m</math>. Multiplying by <math>50</math> on both sides, we get <math>39k=58m</math>.
 
If <math>N</math> is <math>22</math> percent less than one integer <math>k</math>, then <math>N=\frac{78}{100}k=\frac{39}{50}k</math>. In addition, <math>N</math> is <math>16</math> percent greater than another integer <math>m</math>, so <math>N=\frac{116}{100}m=\frac{29}{25}m</math>. Therefore, <math>k</math> is divisible by 50 and <math>m</math> is divisible by 25. Setting these two equal, we have <math>\frac{39}{50}k=\frac{29}{25}m</math>. Multiplying by <math>50</math> on both sides, we get <math>39k=58m</math>.
  
 
The smallest integers <math>k</math> and <math>m</math> that satisfy this are <math>k=1450</math> and <math>m=975</math>, so <math>N=1131</math>. The answer is <math>\boxed{131}</math>.
 
The smallest integers <math>k</math> and <math>m</math> that satisfy this are <math>k=1450</math> and <math>m=975</math>, so <math>N=1131</math>. The answer is <math>\boxed{131}</math>.
 +
 +
==Solution 2==
 +
Continuing from Solution 1, we have <math>N=\frac{39}{50}k</math> and <math>N=\frac{29}{25}m</math>. It follows that <math>k=\frac{50}{39}N</math> and <math>m=\frac{25}{29}N</math>. Both <math>m</math> and <math>k</math> have to be integers, so, in order for that to be true, <math>N</math> has to cancel the denominator of both <math>\frac{50}{39}</math> and <math>\frac{25}{29}</math>. In other words, <math>N</math> is a multiple of both <math>29</math> and <math>39</math>. That makes <math>N=\operatorname{lcm}(29,39)=29\cdot39=1131</math>. The answer is <math>\boxed{131}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2015|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2015|n=II|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:36, 12 May 2021

Problem

Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$.

Solution 1

If $N$ is $22$ percent less than one integer $k$, then $N=\frac{78}{100}k=\frac{39}{50}k$. In addition, $N$ is $16$ percent greater than another integer $m$, so $N=\frac{116}{100}m=\frac{29}{25}m$. Therefore, $k$ is divisible by 50 and $m$ is divisible by 25. Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$. Multiplying by $50$ on both sides, we get $39k=58m$.

The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$, so $N=1131$. The answer is $\boxed{131}$.

Solution 2

Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$. It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$. Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominator of both $\frac{50}{39}$ and $\frac{25}{29}$. In other words, $N$ is a multiple of both $29$ and $39$. That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$. The answer is $\boxed{131}$.

See also

2015 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS