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

Revision as of 20:15, 24 December 2020

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 $1$ and greater than $\frac{41}{42}$ . Thus the only integer the expression can take is $1$ , making the only value for $n$ $42$ . Thus $\boxed{\mathrm{(E)}\ n>84}$

~superagh

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 12: / Line 12: @@
 Since <math>\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}</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>
+<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>.

Page

Toolbox

Search