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if x, y, z,w are all positive integers less than 10 that satisfy the above equation, what is the value of w?

1. x=y-6 2. x+y+z=13

since z > 0, z < 10, and z is divisible by 5 -> z = 5

20 x + 2y + 1 = 10 w + 5

2y + 1 = 5 or 15 -> y = 2 or 7

1) x = y -6 -> y = 7, x = 1, z = 5 and w = 3 -> sufficient 2) x + y + z = 13 -> z = 5, x+y=8 y =2 -> x = 6, w = 12 y = 7 -> x = 1 -> w = 3 -> insufficient

\(\frac{1}{5}y-\frac{2}{5}\) - have to be an integer. It is possible if \(\frac{y}{5}=\frac{2}{5}\) or \(\frac{y}{5}=\frac{7}{5}\)

What is OA? Am I wrong?

note that z must be 5 so that 100x+10y+z is a multiple of 5. since 100x+10y+z=5(10w+5), a multiple of 25, y must be either 2 or 7, as multiples of 25 that do not end in 0 must end in 25 or 75.

1.since alla variables are positive, y must be 7 and x must be 1. suff 2.x+y=8 and as x must be less than 5 for the quotient to have only 2 digits, y must be 7 and x 1. suff

Z can be 5 or 0 since all variables are integers and 100X + 10Y has to be a multiple of 10. Since all variables have to be positive, Z = 5

multiplying both sides by 5 to get rid of fraction we have:

100X + 10Y + Z = 50W + 5Z

Since Z = 5,

100X + 10Y + 5 = 50W + 25

100X + 10Y = 50W + 20

Divide by 10 to reduce

10X + Y = 5W + 2

Get W to one side

(10X + Y - 2) / 5 = W

So Since (10X + Y - 2) is divisible by 5 it follows that Y has to be = 7. This is because 10X will always be divisible by 5 and now you need Y-2 to be divisible by 5, the only integers less than 10 that satisfies that is 7 or 2

Since statement 1 is X = Y - 6 , Y must be greater than 6 because all variables are positive (X has to be positive) , so 7 = Y, then by plugging in X = 1 and since you know Z=5 from before we can find W. W=3

Statement 2, after we find out that Z=5 states

X+Y=8

From the stem we know Y=7 or Y=2

If Y=2 X=6

If we plug that into the stem:

(10X + Y - 2) / 5 = W

60/5 = W

W=12 Since W has to be less than 10, we can conclude that Y=7 and X=1.

Choice: D

Definitely wouldnt be able to do this in 2min though

gmatclubot

Re: integer equation
[#permalink]
18 Feb 2008, 12:39