Tessellate S.T.E.M.S (2019) - Mathematics - Category B - Set 1 - Objective Problem 2

Algebra Level 3

Let a , b , c a, b, c be natural numbers such that 1 a + 1 b + 1 c < 1 \frac{1}{a} +\frac{1}{b}+\frac{1}{c} < 1 . If the maximum value that 1 a + 1 b + 1 c \frac{1}{a} + \frac{1}{b} + \frac{1}{c} can take is p q \frac{p}{q} , where p p and q q are relatively prime positive integers, find p + q p+q .

This problem is a part of Tessellate S.T.E.M.S. (2019)

54 69 83 76

Tessellate S.T.E.M.S. Mathematics by Tessellate S.T.E.M.S. Ma… , 26

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2 solutions

Patrick Corn
Jan 30, 2019

If a , b , c ≥ 3 a,b,c \ge 3 then the maximum attainable value is clearly 1 3 + 1 3 + 1 4 = 11 12 . \frac13 + \frac13 + \frac14 = \frac{11}{12}. The only way to do better is if one of the numbers is 2. 2.

WLOG take a = 2. a=2. Then we have 1 b + 1 c < 1 2 . \frac1{b} + \frac1{c} < \frac12. If b , c ≥ 4 b,c \ge 4 then the maximum attainable value is 1 2 + 1 4 + 1 5 = 19 20 . \frac12 + \frac14 + \frac15 = \frac{19}{20}. The only way to do better is if one of the numbers is 3. 3.

WLOG take b = 3. b=3. In this case we get 1 c < 1 6 , \frac1{c} < \frac16, so the maximum attainable value is when c = 7. c=7. This value is 1 2 + 1 3 + 1 7 = 41 42 , \frac12 + \frac13 + \frac17 = \frac{41}{42}, so the answer is 83 . \fbox{83}.

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Ankush Agarwal
Jan 9, 2019

83=42+41

41=6+14+21

(6+14+21)/42=1/7+1/3+1/2

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