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

 It is currently 04 May 2015, 22:50

### 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

# a and b are two positive integers that are not divisible by

Author Message
TAGS:
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1269
Followers: 23

Kudos [?]: 158 [0], given: 0

a and b are two positive integers that are not divisible by [#permalink]  06 Mar 2007, 00:31
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
a and b are two positive integers that are not divisible by 10, and the sum of the digits of the product of a and b is 1. Which of the following cannot be the remainder when |a-b| is divided by 10?

(I) 5
(II) 9
(III) 0

(A) I only (B) II only (C) III only (D) II and III only (E) I, II and III

Last edited by kevincan on 06 Mar 2007, 10:06, edited 1 time in total.
Intern
Joined: 03 Mar 2007
Posts: 14
Followers: 0

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

Re: PS: Remainder of Special Product [#permalink]  06 Mar 2007, 02:55
According to me the answer should be (B).

a*b = 10^n as the sum of the digits is 1.
=> a*b = (5^n) * (2^n)
=> |a-b| is odd but not divisible by 5.

is the answer and explanation correct?

~newbie here...
Manager
Joined: 09 Jan 2007
Posts: 240
Followers: 1

Kudos [?]: 7 [0], given: 0

There are three numbers 1, 10, 100 can be product of a and b, and the sum of digits would be 1. Beyond 100 one of the numbers has to be divisible by 10.

1 = 1*1
10 = 2*5
100= 4*25

|1-1| = 0, remainder=0, when divided by 10
|2-5| = 3, remainder=3, when divided by 10
|4-25| = 21, remainder=1, when divided by 10

so the answer is (I) and (II). But do not see the choice with (I) and (II). Am I doing something wrong?
Senior Manager
Joined: 29 Jan 2007
Posts: 450
Location: Earth
Followers: 2

Kudos [?]: 40 [0], given: 0

I go with B.

rdg,
You can try 100 = 5* 20 and then rule out 1.
I solved this similar to how you did it.
Director
Joined: 14 Jan 2007
Posts: 780
Followers: 2

Kudos [?]: 71 [0], given: 0

the product of a and b would be :

1 = 1*1 remainder is zero
10= 5*2 remainder is 3
100 = 25*4 remainder is 1
1000 = 125*8 remainder is 7
10000 = 625*16 remainder is 9
100000 = 3125*32 remainder is 3
and so on......

'A' seems to be the answer.
Manager
Joined: 09 Jan 2007
Posts: 240
Followers: 1

Kudos [?]: 7 [0], given: 0

Thanks. I should have considered with more numbers.
Senior Manager
Joined: 29 Jan 2007
Posts: 450
Location: Earth
Followers: 2

Kudos [?]: 40 [0], given: 0

Kevincan question seems to be wrong then...as all three can be remainder.
And there is no option that says 'none'
Similar topics Replies Last post
Similar
Topics:
If a and b are positive integers, is (10^a) + b divisible by 2 29 Jul 2009, 16:50
If a and b are positive integers divisible by 6, is 6 the 3 05 Aug 2008, 19:21
1 If a and b are positive integers divisible by 6, is 6 the 1 18 Feb 2008, 10:57
If a and b are positive integers divisible by 6, is 6 the 5 17 Jun 2007, 00:35
a and b are two positive integers that are not divisible by 1 24 Sep 2006, 14:03
Display posts from previous: Sort by