Positive integers x, y and z, where 0

Question

Positive integers x, y and z, where 0<x<100 and 0<y<100, are such that:

\frac{x+y}{x} = z^3

What is the largest possible value of x+y+z?

A. 90
B. 98
C. 100
D. 106
E. 114

IanStewart · Answer

(x + y)/x = x/x + y/x = 1 + (y/x)

and since y/x < y < 100, we know z^3 is less than 100. So z can only be 2, 3 or 4. If we make z = 2, we get

(x + y)/x = 8
x + y =8x
y = 7x

and since y < 100, the maximum possible value of x is 14. In that case, y = 98, and z = 2, so x + y + z = 114. Since there is no greater answer choice, there's no reason to test other values of z.

Fdambro294 · Answer

0 < X < 100 --- and --- 0 < Y < 100

there is No Restriction at the outset of the question on Z, except that it must = (+)Positive Integer because the Result of X + Y / X = (Z)^3 = Perfect Cube, which must be an Integer CUBED

1st) Re-arrange the Equation

(X + Y)/X = (Z)^3

(X/X) + (Y/X) = (Z)^3

(Y / X) + 1 = PERFECT CUBE

2nd) We need a (+)Integer b/w 0 and 100 to divide into another (+)Integer b/w 0 and 100 and produce an Integer Result that is -1 SHORT of a Perfect Cube.

Since we want to Maximize --- (X + Y + Z) --- we want to be able to create a Divisor X that is LARGE but Still can divide into a LARGE Dividend Y to produce an Integer -1 Short of a Perfect Cube.

Furthermore, the MAXIMUM Integer quotient we can create when we divide Y/X would be:

99/1 = 99

Adding +1 to 99 --- the Perfect Cube (Z)^3 must be a Perfect Cube LESS THAN< 100

If we aim for a higher Perfect Cube such as 64, since Y must be between 0 and 100, we can only have the following situation:

Y/X + 1 = Perfect Cube (Z)^3

63 / 1 + 1 = 64

Y + X + Z = 63 + 1 + 4 = 67

Because of this, we are better off aiming for a SMALLER Perfect Cube in order to create a Larger Divisor X.

Aiming for Perfect Cube = 8

Y/X must = 7

This means the Dividend Y must be a Multiple of 7 if it produces a Quotient of 7.

The Largest Multiple of 7 LESS THAN< 100 would be 98

Let Y = 98

98/X = 7 ----- which means X must = 14

and finally,

Y/X + 1 = (Z)^3

98/14 + 1 =

7 + 1 = 8

8 = (Z)^3 ---- taking the ODD Root of (Z)^3 gives us: Z = 2

Solution:

Y = 98

X = 14

Z = 2

98 + 2 + 14 = 114, the largest Answer Choice Available.

Answer -E-

bumpbot · Answer

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Positive integers x, y and z, where 0<x<100 and 0<y<100, are such that

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