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

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 , hence 
  but  does not divides ,
                                                                                                      -(1)

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

  but ,
                                                                                                      -(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

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

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

@@ Line 1: / Line 1: @@
-#REDIRECT [[2002 AMC 12B Problems/Problem 5]]
+{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #4]] and [[2002 AMC 10B Problems|2002 AMC 10B #7]]}}
+== Problem ==
+Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the following statements is '''not ''' true:
+<math>\mathrm{(A)}\ 2\ \text{divides\ }n
+\qquad\mathrm{(B)}\ 3\ \text{divides\ }n
+\qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n
+\qquad\mathrm{(D)}\ 7\ \text{divides\ }n
+\qquad\mathrm{(E)}\ n > 84</math>
+== Solution 1==
+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>
+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>.
+== 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 ==
+{{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}}
+{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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