Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 16:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# 100x+10y+z)/5=10w+z if x, y, z,w are all positive integers

Author Message
TAGS:
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

100x+10y+z)/5=10w+z if x, y, z,w are all positive integers [#permalink]  25 Jan 2008, 09:55
1
This post was
BOOKMARKED
(100x+10y+z)/5=10w+z

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

Last edited by marcodonzelli on 25 Jan 2008, 10:27, edited 1 time in total.
Senior Manager
Joined: 22 Sep 2005
Posts: 279
Followers: 1

Kudos [?]: 89 [1] , given: 1

Re: integer equation [#permalink]  25 Jan 2008, 10:21
1
KUDOS
marcodonzelli wrote:
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

E.

1) NOT SUF because z and w can be any positive integer <10
2) NOT SUF. We can have the following solutions:

X Y Z
2 3 8
3 4 6
5 6 2

1 and 2 together:

From x=y-6, the only possible solution is that:
y=8, x=2 and z=3
But w can be any other positive number<10, such as 4,5....

So E.
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

Re: integer equation [#permalink]  25 Jan 2008, 10:28
netcaesar wrote:
marcodonzelli wrote:
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

E.

1) NOT SUF because z and w can be any positive integer <10
2) NOT SUF. We can have the following solutions:

X Y Z
2 3 8
3 4 6
5 6 2

1 and 2 together:

From x=y-6, the only possible solution is that:
y=8, x=2 and z=3
But w can be any other positive number<10, such as 4,5....

So E.

I am sorry, I missed to write down the equation. look at the Q now
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [1] , given: 359

Re: integer equation [#permalink]  25 Jan 2008, 12:48
1
KUDOS
Expert's post
A

$$10w+z=\frac{100x+10y+z}{5}$$

100x and 10y are divisible by 5. Therefore, z is also divisible 5 and equal z=5

$$10w+5=\frac{100x+10y+5}{5}$$

$$w=2x+\frac{1}{5}y-\frac{2}{5}$$

y have to be y=2 or y=7.

1. x=y-6. Only y=7 satisfies this condition. sufficient.

2. x+y+5=13 ==> x+y=8. both y=2 and y=7 satisfy condition. insufficient.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

Re: integer equation [#permalink]  08 Feb 2008, 07:30
walker wrote:
A

$$10w+z=\frac{100x+10y+z}{5}$$

100x and 10y are divisible by 5. Therefore, z is also divisible 5 and equal z=5

$$10w+5=\frac{100x+10y+5}{5}$$

$$w=2x+\frac{1}{5}y-\frac{2}{5}$$

y have to be y=2 or y=7.

1. x=y-6. Only y=7 satisfies this condition. sufficient.

2. x+y+5=13 ==> x+y=8. both y=2 and y=7 satisfy condition. insufficient.

are you sure y can be 2?
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [0], given: 359

Re: integer equation [#permalink]  08 Feb 2008, 08:21
Expert's post
marcodonzelli wrote:
are you sure y can be 2?

$$w=2x+\frac{1}{5}y-\frac{2}{5}$$

$$\frac{1}{5}y-\frac{2}{5}$$ - 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?
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 629
Followers: 3

Kudos [?]: 139 [0], given: 1

Re: integer equation [#permalink]  08 Feb 2008, 10:27
marcodonzelli wrote:
(100x+10y+z)/5=10w+z

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

VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

Re: integer equation [#permalink]  17 Feb 2008, 22:25
walker wrote:

$$w=2x+\frac{1}{5}y-\frac{2}{5}$$

$$\frac{1}{5}y-\frac{2}{5}$$ - 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
VP
Joined: 22 Oct 2006
Posts: 1443
Schools: Chicago Booth '11
Followers: 8

Kudos [?]: 154 [0], given: 12

Re: integer equation [#permalink]  18 Feb 2008, 11:39
(100X + 10Y + Z ) / 5 = 10W + Z

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
Re: integer equation   [#permalink] 18 Feb 2008, 11:39
Similar topics Replies Last post
Similar
Topics:
If x, y, and z are all positive integers, is x + y + z even? 1 22 Oct 2013, 07:07
7 If x, y, and z are all positive integers, what is the 21 07 Nov 2012, 13:53
6 If x and y are positive integers and 7 21 Feb 2012, 13:45
1 If x, n, and y are all positive integers, is x^n divisible 4 25 Jan 2011, 20:42
V, W, X, Y, and Z are all positive integers. W is greater 1 26 Nov 2005, 06:40
Display posts from previous: Sort by