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Bunuel
If \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{x}\), which of the following must be an integer?


I. \(\frac{x}{8}\)

II. \(\frac{x}{12}\)

III. \(\frac{x}{24}\)


A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III

1/2+1/3+1/4=13/12. Thus x=12.

Not just put 12 in place of x in all three options we get only 2 II to be an integer.

Option B.
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If the question used "COULD be true" instead of "must be true", would the answer be D?

Because we know that \(\frac{13}{12}\) = \(\frac{13}{x}\)

So x could also be a multiple of 12, therefore \(\frac{x}{24}\) could also be true.

Am I correct?
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1/2 + 1/3 + 1/4 = 13/x

=>13/12 = 13/x
=>13x = 156
=>x = 156/13
=>x=12

A. 12/8 = 3/2 - Not an Integer
B. 12/12 = 1 - Integer
C. 12/24 = 1/2 - Not an integer

Hence B is the answer
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Bunuel
If \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{x}\), which of the following must be an integer?


I. \(\frac{x}{8}\)

II. \(\frac{x}{12}\)

III. \(\frac{x}{24}\)


A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III

Multiplying by 12x we have:

6x + 4x + 3x = 156

13x = 156

x = 12; thus, x/12 must be an integer.

Answer: B
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Bunuel
If \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{x}\), which of the following must be an integer?


I. \(\frac{x}{8}\)

II. \(\frac{x}{12}\)

III. \(\frac{x}{24}\)


A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III

Given: \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{x}\)

Asked: Which of the following must be an integer?

\(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{x}\)

\(\frac{(6+4+3)}{12} = \frac{13}{x}\)

\(\frac{13}{12} = \frac{13}{x}\)

\(\frac{1}{12} = \frac{1}{x}\)

x = 12

I. \(\frac{x}{8}\)

\(\frac{12}{8} = \frac{3}{2} = 1.5\)

NOT AN INTEGER

II. \(\frac{x}{12}\)

\(\frac{12}{12} = 1\)

INTEGER

III. \(\frac{x}{24}\)

\(\frac{12}{24} = \frac{1}{2} = 0.5\)

NOT AN INTEGER


IMO B
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