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| Getting Fruity With 73 And 2005 (Posted on 2007-08-06) |
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Harold and Jonah each has some oranges, apples, and pears. Each of them has 73 pieces of fruit. Curiously, if for each of them you sum the squares of the number of oranges, apples, and pears, you get 2005 in both cases.
If Harold has six fewer apples than Jonah, and each man has fewer oranges compared to other fruits, what do they have each?
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Submitted by K Sengupta
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Rating: 3.5000 (2 votes)
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Solution:
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Harold has 12 oranges, 30 apples and 31 pears, while Jonah has 15 oranges, 36 apples and 22 pears.
EXPLANATION:
Let us denote the respective number of oranges, apples and pears as R, A and P. Then, by the conditions of the problem:
R+A+P = 73; R^2 + A^2 + P^2 = 2005
At the outset, we can assume that R ≤ A ≤ P, since the other solutions will be a permutation of this.
Then, 3R ≤ R+A+P = 73 < 75
Or, R< 25
Again, P+A = 73 – R, ...(*); and:
P^2 + A^2 = 2005 – R^2, giving:
S^2 = 2(2005 – R^2) – (73-R)^2 = -1319 + 146*R -3*R^2, where:
S = P-A.
We now display the choices (R, S^2) for R< 25. Since S^2< 0 for R< 12, only the values R ≥ 12 are shown here.
(R, S^2) = (12, 1); (13, 72); (14, 137); (15, 196); (16, 249); (17, 296); (18, 337); (19, 372); (20, 401); (21, 424); (22, 441); (23, 452); (24, 457)
From the foregoing, we observe that, the values of S^2 is a perfect square whenever (P-A)^2 = S^2 = 1, 196, 441 corresponding to R = 12, 15, 22.
Recalling that R ≤ A ≤ P, we observe that:
R = 12 gives (P-A, P+A) = (1, 61)(from (*)), so that:
(P, A) =(31, 30)
R = 15 gives (P-A, P+A) = (14, 58), so that: (P, A) = (36, 22)
R = 22 gives (P-A, P+A) = (21, 51), so that (P, A) = (36, 15) , which contradicts the restriction R< P.
Accordingly, without the restriction R < A < P, it follows that:
(R, A, P) (- (12, 30, 31), (15, 22, 36) (in some order)
By the problem, R< A and R< P, and so:
(R, A, P) = (12, 30, 31); (12, 31, 30); (15, 22, 36); (15, 36, 22)
Since Harold has precisely 6 fewer apples than Jonah, it now follows that:
For Harold; (R, A, P) = (12, 30, 31), and:
For Jonah; (R, A, P) = (15, 36, 22)
Consequently, Harold has 12 oranges, 30 apples and 31 pears, while Jonah has 15 oranges, 36 apples and 22 pears.
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Also refer to the computer aided solution posted by Charlie in this location .
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