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Question

Another problem which killed me on time... can solve it but not fast enough. Any ideal solutions for this?

sgrover · Answer

75 = 49+25+1
Sum = 7 + 5 + 1

(13)

Not sure if there is any method for this.. worked by picking no.s..

MA · Answer

as for the quickest approach:

first laid down all squares below 75 as under:
1, 4, 9, 25, 36, 49, 64

then deduct one of the largest square (lets say 36) from 75. we get 41. none of the remaining two squares make 41, if we add them up..

now go with 49. 75-49 = 26. we can quickly visualize 1, which if taken out from 26 remains 25 which is a square in our list.

so the desired squares are 1, 25, 49. sum of the sqrt of these squares (1, 5, 7) makes up 13.

ywilfred · Answer

25 + 49 + 1 = 75

Sum of integers = 5 + 7 + 1 = 13

Ans E

Zooroopa · Answer

Agree with E.

Point to note: 1,4,9,16,25 cannot form 75 on their own. They need to be with the others. So start from the bottom and get your combination.

ps_dahiya · Answer

I noted down the last digits of the squares: 1, 4,9,16,25,36,49 that will be 
1,4,9,6,5,6,9
If we select 3 what sum can result in last digit as 5.
Only these are possible:
4,5,6 - 4 + 25 + 16 or 4 + 25 + 36 ----NO
1,5,9 - 1+25 + 9 ---NO or 1 + 25 + 49 - YES

GMATT73 · Answer

Used the same method.

(E)

Not sure if it`s the quickest, but it only took about 45 seconds.

gmatmathguru · Answer

Given (x^2+y^2+z^2) = 75, what is x+y+z

1. Note that (x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)

which means (x+y+z)^2 = 75 + 2(xy+yz+zx)

because 2(xy+yz+zx) is even and 75 is odd, the sum (x+y+z) should be odd, hence the choices 14 and 16 are wrong.

2. Note that the problem has a trivial solution when x = y = z = 5.

The sum = x+y+z = 15 which is the maximum value of the sum (x+y+z).

hence, the sum x+y+z <= 15.

from 1 and 2, and the fact that x,y,z are distinct positive integers (and hence the integers cannot be x=y=z=5, and the sum cannot be 15) the sum is 13.

The connection to Geometry:

Note that sqrt(x^2+y^2+z^2) represents the longest diagonal of a rectangular solid with sides x, y and z. Given the diagonal, the sum of the sides x+y+z is maximum when the rectangular solid is a cube i.e. x = y = z.

-mathguru

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