Tough DS: Factors & Multiples

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ash g: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Mon Oct 15, 2007 11:47 pm; Thanked: 1 times

by ash g » Thu Mar 20, 2008 6:03 pm

hi all,

Really need to hack this one. Where did I go wrong on this one ?
DS Factors & Multiples problems are usually hard to crack. So really wanna understand whats happening below.

Source: MGMAT2
Level: 700-800

Q. When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

Their answer: E
My Answer: C

Their Explanation : Available if needed
My Explanation : Below

S1: When x/3 is divided by 2, the remainder is 1.

Which is another way of saying that x/6 leaves a remainder of 1(Is this correct). So...

Multiples of 6 = 0 6 12 18 24 30 36
x can be = 1 7 13 19 25 31 37
Remainder of x/4 = 1 3 1 3 1 3 1

So Insufficient

S2: x is divisible by 5
x can be = 0 5 10 15 20 25 30 35
Remainder of x/4 = 0 1 2 3 0 1 2 3

S12:
Overlap of S1 and S2 = 25,60,....
Remainder of x/4 = 1

Why is [C] not the answer?

Regards,
Ash

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Source: Beat The GMAT — Data Sufficiency | Thu Mar 20, 2008 6:03 pm

kashman_x7: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sun Jan 06, 2008 2:08 pm

by kashman_x7 » Thu Mar 20, 2008 10:14 pm

You are nearly correct except when you combine the two statements

x = 25, 55, 85 etc
which implies when x/4 remainder = 1,3,1,3 etc so E is the correct answer

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mmukher: Senior | Next Rank: 100 Posts; Posts: 34; Joined: Wed Feb 20, 2008 7:15 pmFollowed by: 1 member; Thanked: 6 times

by mmukher » Sun Mar 23, 2008 5:15 am

Aah, so it could be 1 OR 3 so we aren't sure and the answer is E.
That was tricky, thanks!

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humeixia: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Thu Apr 10, 2008 1:20 pm; Thanked: 1 times

by humeixia » Thu Apr 17, 2008 6:48 am

Any deep analysis for this question?

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Sat May 09, 2009 11:08 am

Ian please help.

I am getting as below:

x/3 = 2K+1 so x = 6k+3 for k=1, 3, 5 etc 4 gives remainder of 1, for 2/4/6 etc gives 3 as 6k will be divisible by 4

from x as a multiple of 5: 15(k=2) 45(k=7) so not sufficient

combine also does not help

so E

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mmslf75: Legendary Member; Posts: 503; Joined: Sun Aug 09, 2009 9:53 pmFollowed by: 2 members; Thanked: 31 times

by mmslf75 » Mon Aug 10, 2009 8:31 am

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vikram_k51: Master | Next Rank: 500 Posts; Posts: 208; Joined: Sat Jan 31, 2009 11:32 am; Location: Mumbai; Thanked: 2 times

by vikram_k51 » Mon Aug 10, 2009 8:58 am

Q. When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

The question can be rephrased as "Is x=4k+3?"

From 1 x= 6k+1 Or 4k+2k+1 Not sufficient

From 2 x=5k2 Not Sufficient X can be 5,10,15 In each case the rem when x is divided by 4 is different.

Combining the 2 we have x=5k+k+1 Not Suff

Hence E

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cubicle_bound_misfit: Master | Next Rank: 500 Posts; Posts: 246; Joined: Mon May 19, 2008 7:34 am; Location: Texaco Gas Station; Thanked: 7 times

by cubicle_bound_misfit » Mon Aug 10, 2009 10:35 am

Hope this one will help.
So basically you have to find a number which is of the form

number == multiple of 6 +1 == multiple of 5

now multiple of 6 has a pattern when it comes to unit digit

6 2 8 4 0 | 6 2 8 4 0| 6 2 ...

the only way you can get a multiple of 5 from this pattern is wen you add 1 to 4 th term of every series ie 24 + 1 ( 4th term of 1st series)
54+1 ( 4th term of 2nd series)

lets see some of those multiples

24 + 1 = 25
54+1 = 55
84+1 = 85
now divide them with 4. you will get different remainder.

Ian, if there is a better method, please help us.
Thanks.
CBM

Cubicle Bound Misfit

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Mon Aug 10, 2009 8:03 pm

IMO E
question asks whether the number is of the form 4k+3

1) x/3= 2k+1
x=3(2k+1)
number is an odd multiple of 3
3,9,..
not SUFF

2)X=5k=4k+k.
k can be 0,1,2,3,4. not suff
combined
take 15
take 45

NOT SUFFICIENT

algebraically,
number should be 3*5 *(2k+1)
2k+1 can give you remainder 1 or 3 when divided by 4
so not SUFF

The powers of two are bloody impolite!!

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ghacker: Master | Next Rank: 500 Posts; Posts: 148; Joined: Wed Jun 03, 2009 8:04 pmFollowed by: 1 member; Thanked: 18 times

by ghacker » Mon Aug 10, 2009 9:29 pm

The answer is E

(1) When x/3 is divided by 2, the remainder is 1.: This means that x/3 is odd so X must be an odd multiple of 3 ( assuming that x/3 is an int ....) so this is insufficient , Y ? if x =15 when we divide 15 by 4 we get remainder 3
also x can be 45 , in this situation also we get remainder 3

(2) x is divisible by 5.

This means x is a multiple of 5 .... clearly insufficient

Take 1 and 2 together , then still we will get the 15 and 45 issue so

The answer is E

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mark412: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Mon Jul 27, 2009 5:46 pm; Location: LA,CA

Answer is E

by mark412 » Mon Aug 17, 2009 3:42 pm

Q. When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

(1) I resimplified this to read x/6 (bc x/3/2 is really x/3 x 1/2). then from there I plugged in possible values for x. You will find some answers like 7 will hold true while 13 wont.

insuff.

(2) Again, plug in choices. 5 & 10 wont work but 15 will.

insuff.

Answer is E.

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