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Difference between revisions of "2002 AMC 12B Problems/Problem 4"

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<math>\mathrm{(A)}\ 2\ \text{divides\ }n
 
<math>\mathrm{(A)}\ 2\ \text{divides\ }n
 
\qquad\mathrm{(B)}\ 3\ \text{divides\ }n
 
\qquad\mathrm{(B)}\ 3\ \text{divides\ }n
−
\qquad\mathrm{(C)}\ 6\ \text{divides\ }n  
+
\qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n  
 
\qquad\mathrm{(D)}\ 7\ \text{divides\ }n
 
\qquad\mathrm{(D)}\ 7\ \text{divides\ }n
 
\qquad\mathrm{(E)}\ n > 84</math>
 
\qquad\mathrm{(E)}\ n > 84</math>
  
−
== Solution ==
+
== Solution 1==
−
Since <math>\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}</math>,
+
Since <math>\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}</math>, <math>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</math>
  
−
<cmath>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</cmath>
+
From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. The only answer choice that is not true is <math>\boxed{\mathrm{(E)}\ n>84}</math>.
  
−
From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. Thus the answer is <math>\boxed{\mathrm{(E)}\ n>84}</math>.
+
== Solution 2 (no limits)==
 +
Since <math>\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}</math>, it is very clear that <math>n=42</math> makes the expression an integer*. Because <math>n</math> is a positive integer, <math>\frac{1}{n}</math> must be less than or equal to <math>1</math>. Thus, the only integer the expression can take is <math>1</math>, which means the only value for <math>n</math> is <math>42</math>. Thus <math>\boxed{\mathrm{(E)}\ n>84}</math>
 +
 
 +
~superagh
 +
 
 +
== Solution 2.1 (faster ending to solution 2) ==
 +
Once you find <math>n=42</math>, you can skip the rest of Solution 2 by noting that <math>n=42=2\times3\times7</math>, which satisfies A, B, C, D, but not E. Thus <math>\boxed{\mathrm{(E)}\ n>84}</math> must be the answer by PoE (Process of Elimination).
 +
 
 +
 
 +
~speed tip by rawr3507
 +
 
 +
== Solution 3(similiar to solution2)==
 +
Cross multiplying and adding the fraction we get the fraction to be equal to <math>\frac {41n + 42}{42n}</math>, This value has to be an integer. This implies,
 +
 +
<math>42n|(41n+42)</math>.
 +
 
 +
=> <math>42|(41n + 42)</math>
 +
  but <math>42|42</math>, hence <math>42|41n</math>
 +
  but <math>42</math> does not divides <math>41</math>,
 +
  <math>42|n</math>                                                                                                    -(1)
 +
 
 +
=> <math>n|(41n+42)</math>
 +
  but <math>n|41n</math>,
 +
  <math>n|42</math>                                                                                                    -(2)
 +
 
 +
from (1) and (2) we get that n=42.
 +
Comparing this with the options, we see that option E is the incorrect statement and hence E is the answer.
 +
 
 +
~rudolf1279
  
 
== See also ==
 
== See also ==
Line 21: Line 49:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 04:34, 9 November 2024

The following problem is from both the 2002 AMC 12B #4 and 2002 AMC 10B #7, so both problems redirect to this page.

Problem

Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:

$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$

Solution 1

Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$, $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$

From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$. The only answer choice that is not true is $\boxed{\mathrm{(E)}\ n>84}$.

Solution 2 (no limits)

Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$, it is very clear that $n=42$ makes the expression an integer*. Because $n$ is a positive integer, $\frac{1}{n}$ must be less than or equal to $1$. Thus, the only integer the expression can take is $1$, which means the only value for $n$ is $42$. Thus $\boxed{\mathrm{(E)}\ n>84}$

~superagh

Solution 2.1 (faster ending to solution 2)

Once you find $n=42$, you can skip the rest of Solution 2 by noting that $n=42=2\times3\times7$, which satisfies A, B, C, D, but not E. Thus $\boxed{\mathrm{(E)}\ n>84}$ must be the answer by PoE (Process of Elimination).


~speed tip by rawr3507

Solution 3(similiar to solution2)

Cross multiplying and adding the fraction we get the fraction to be equal to $\frac {41n + 42}{42n}$, This value has to be an integer. This implies,

$42n|(41n+42)$.

=> $42|(41n + 42)$ 

  but $42|42$, hence $42|41n$
  but $42$ does not divides $41$,
  $42|n$                                                                                                    -(1)

=> $n|(41n+42)$ 

  but $n|41n$,
  $n|42$                                                                                                    -(2)

from (1) and (2) we get that n=42. Comparing this with the options, we see that option E is the incorrect statement and hence E is the answer.

~rudolf1279

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AMC 10 Problems and Solutions
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AMC 12 Problems and Solutions

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