2002 AMC 12B Problems/Problem 4

Revision as of 18:26, 19 August 2020 by Superagh (talk | contribs) (Solution 2 (no limits))
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

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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