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Algebra Difficulty 2.2 Junior Find the answer

When three positive integers are added in pairs, the resulting sums are 998, 1050, and 1234. What is the difference between the largest and smallest of the three original positive integers?

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

Solution

Suppose that the three integers are x,yx, y and zz where x+y=998x+y=998, x+z=1050x+z=1050, and y+z=1234y+z=1234. From the first two equations, (x+z)(x+y)=1050998(x+z)-(x+y)=1050-998 or zy=52z-y=52. Since z+y=1234z+y=1234 and zy=52z-y=52, then (z+y)+(zy)=1234+52(z+y)+(z-y)=1234+52 or 2z=12862z=1286 and so z=643z=643. Since z=643z=643 and zy=52z-y=52, then y=z52=64352=591y=z-52=643-52=591. Since x+y=998x+y=998 and y=591y=591, then x=998y=998591=407x=998-y=998-591=407. The three original numbers are 407, 591, and 643. The difference between the largest and smallest of these integers is 643407=236643-407=236.

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