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Difference between revisions of "1996 AJHSME Problems/Problem 4"

(Created page with "==Problem== <math>\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}=</math> <math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{2}{3} \qquad \text{(C)}\ \dfrac{3}{2} \qquad \text{...")
 
 
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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Latest revision as of 00:24, 5 July 2013

Problem

$\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}=$

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{2}{3} \qquad \text{(C)}\ \dfrac{3}{2} \qquad \text{(D)}\ \dfrac{17}{3} \qquad \text{(E)}\ \dfrac{34}{3}$

Solution 1

First, notice that each number in the numerator is a multiple of $2$, and each number in the denominator is a multiple of $3$. This suggests that each expression can be factored. Factoring gives:

$\dfrac{2(1+2+3+\cdots + 17)}{3(1+2+3+\cdots + 17)}$

Since all the material in the parentheses is the same, the common factor in the numerator and the denominator may be cancelled, leaving $\frac{2}{3}$, which is option $\boxed{B}$.

Solution 2

There are $17$ terms in the numerator $2 + 4 + 6 + ... + 30 + 32+ 34$. Consider adding those terms, but in a different order. Start with the last two terms, $2 + 34$, and then add the next two terms on the outside, $4 + 32$, and continue. You will get $8$ pairs of numbers that add to $36$, while the $9^{th}$ number in the middle will be alone. That number is $18$. Adding all the numbers gives $8\cdot 36 + 18 = 306$.

Similarly, the denominator has $17$ terms of $3 + 6 + 9 + ... + 45 + 48 + 51$. There are $8$ pairs of numbers that add up to $54$, with the $9^{th}$ number in the center being $27$. The total of all the numbers is $8\cdot 54 + 27 = 459$.

The answer is $\frac{306}{459}$. Eyeballing the options, the fraction is clearly under $1$, but more than $\frac{1}{2}$. Thus, the answer must be $\frac{2}{3}$, or $\boxed{B}$. Alternately, you can do the work by factoring out a $9$ in the numerator to give $\frac{34}{51}$. Factoring out $17$ will give the desired answer.

Solution 3

Start by finding a pattern:

$\frac{2}{3} = \frac{2}{3}$

$\frac{2 + 4}{3 + 6} = \frac{6}{9} = \frac{2}{3}$

$\frac{2 + 4 + 6}{3 + 6 + 9} = \frac{12}{18} = \frac{2}{3}$

Each step doesn't seem to change the value of the fraction, so $\boxed{B}$ is the right answer.

See also

1996 AJHSME (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 AJHSME/AMC 8 Problems and Solutions

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