Maths Olympiad Prep

Track / Stage 4 / 291 of 340 #551 of 1964

Problem 551

AMC 12 late, AIME early
Algebra Difficulty 5.0 Find the answer

What is the largest integer that can be placed in the box so that 11<23\frac{\square}{11}<\frac{2}{3} ?

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A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Since 33 is positive, 11<23\frac{\square}{11}<\frac{2}{3} implies

33(11)<33(23) 33\left(\frac{\square}{11}\right)<33\left(\frac{2}{3}\right)

which simplifies to 3×<223 \times \square<22.

The largest multiple of 3 that is less than 22 is 3×73 \times 7, so this means the number in the box cannot be larger than 7 .

Indeed, 711=2133\frac{7}{11}=\frac{21}{33} is less than 23=2233\frac{2}{3}=\frac{22}{33}, so the answer is 7 .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.