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A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts

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A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 28 Apr 2016, 01:12
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A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12

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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 28 Apr 2016, 02:21
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Bunuel wrote:
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12


A good Q..

the equation that can be formed for n peanuts = > n=9x+6..
the equation that can be formed for m peanuts = > m=12y+4..


so mn = (9x+6)(12y+4) where both x and y are integers..
\((9x+6)(12y+4) = 3(3x+2)*4(3y+1) = 12(9xy+6y+3x+2) = 12*3(3xy+2y+x) +12*2.\).
Now 12*3(3xy+2y+x) is Div by 18 hence remainder would be 24 div by 18..
so R =24-18 = 6
C
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 29 Apr 2016, 00:48
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Bunuel wrote:
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12


n = 9x + 6
m = 12y + 4

nm = (9x + 6)*(12y + 4) = 108xy + 36x + 72y + 24
Remainder of nm/18 = (108xy + 36x + 72y + 24)/18

Observe that the first three terms are a multiple of 18
24 when divided by 18 leaves remainder 6
Hence mn/18 will leave remainder 6

Correct Option: C
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 29 Apr 2016, 08:17
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n = 9x + 6
m = 12y + 4

nm = (9x + 6)*(12y + 4) = 108xy + 36x + 72y + 24
Remainder of nm/18 = (108xy + 36x + 72y + 24)/18
The first three terms are a multiple of 18
24 when divided by 18 leaves remainder 6
Hence, mn/18 will leave remainder 6

Option :C
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 11 Jul 2017, 20:56
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Bunuel wrote:
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12


i multiplied the 2 remainder 6 & 4 and divided the product by 18 (6x4/18). the remainder was 6.
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 19 Jul 2017, 04:01
How I did that
9(x)+6=15
12(x)+4=16
Assume they are actually 15 and 16 take unit digit 5*6=30
30/8=24
remainder 6
Is it right approach
Expert opinion please
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 20 Aug 2017, 23:28
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Easy way
n=9q+6 or n can be 6, 15, 24....
m=12q+4 or n can be 4, 16, 28...
take any 2 nos from above series
NM=6X4=24
24/18=reminder=6.
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 17 Sep 2017, 04:25
gps5441 wrote:
Easy way
n=9q+6 or n can be 6, 15, 24....
m=12q+4 or n can be 4, 16, 28...
take any 2 nos from above series
NM=6X4=24
24/18=reminder=6.


hi

how can you say so ....?

take 15 and 16 for example ...
nm = 15 x 16

now (15 x 16)/18 leaves remainder 1
and yes, obviously, it will work with 6, and 4, as the numbers themselves can be 6, and 4 ...

if you have anything to say, please say to me ...

thanks in advance ..
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A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post Updated on: 01 Oct 2018, 21:19
Bunuel wrote:
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12


assuming one peanut per smaller bag,
least possible value of n=9+6=15
and least possible value of m=12+4=16
nm=15*16=240
240/18 leaves a remainder of 6
C

Originally posted by gracie on 17 Sep 2017, 11:46.
Last edited by gracie on 01 Oct 2018, 21:19, edited 1 time in total.
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 21 Sep 2017, 18:49
gmatcracker2017 wrote:
gps5441 wrote:
Easy way
n=9q+6 or n can be 6, 15, 24....
m=12q+4 or n can be 4, 16, 28...
take any 2 nos from above series
NM=6X4=24
24/18=reminder=6.


hi

how can you say so ....?

take 15 and 16 for example ...
nm = 15 x 16

now (15 x 16)/18 leaves remainder 1
and yes, obviously, it will work with 6, and 4, as the numbers themselves can be 6, and 4 ...

if you have anything to say, please say to me ...

thanks in advance ..


15x16=240
240/18 leaves a reminder of 6.

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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 25 Aug 2018, 21:23
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chetan2u niks18 Bunuel VeritasKarishma GMATPrepNow gmatbusters

How about this approach?

Bag 1: 9k+ 6
Bag 2: 12m + 4

Multiplying peanuts of bag 1 and bag 2:
3(3K+2) * 4 (3m+1)
or
12 (3k+2) (3m+1) . . . . . (1)

Q asks: remainder when divisor is 18

can I use concept LCM of 12 and 18 is 36 beyond (1) ?
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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 25 Aug 2018, 21:30
1
adkikani wrote:
chetan2u niks18 Bunuel VeritasKarishma GMATPrepNow gmatbusters

How about this approach?

Bag 1: 9k+ 6
Bag 2: 12m + 4

Multiplying peanuts of bag 1 and bag 2:
3(3K+2) * 4 (3m+1)
or
12 (3k+2) (3m+1) . . . . . (1)

Q asks: remainder when divisor is 18

can I use concept LCM of 12 and 18 is 36 beyond (1) ?


12(3k+1)(3m+2)..
Next step should be 12(9mk+3m+6k+2)=(12*9mk+12*3m+12*6k+12*2)
All terms except 12*2 are divisible by 18...
So 24 divided by 18 leaves a remainder of 6.
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 30 Sep 2018, 20:43
chetan2u wrote:
Bunuel wrote:
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

A. 3
B. 5
C. 6
D. 10
E. 12


A good Q..

the equation that can be formed for n peanuts = > n=9x+6..
the equation that can be formed for m peanuts = > m=12y+4..


so mn = (9x+6)(12y+4) where both x and y are integers..
\((9x+6)(12y+4) = 3(3x+2)*4(3y+1) = 12(9xy+6y+3x+2) = 12*3(3xy+2y+x) +12*2.\).
Now 12*3(3xy+2y+x) is Div by 18 hence remainder would be 24 div by 18..
so R =24-18 = 6
C


picking numbers is so much easier for this.....any particular reason why we do the algebra?
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A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts  [#permalink]

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New post 01 Oct 2018, 09:21
What I did is simply expressed the statement in the form of N = QK + R

n = 9x + 6
=> 2n = 18x + 12 (multiplying by 2 on both sides) -- i)

m = 12y + 4
(3/2)m = 18y +6 (multiplying by 3/2 on both sides) --ii)


we know if r1 and r2 are the two reminders of n and m when divided by a, then if mn is divided by a; the remainder will be r1*r2

Multiplying i) and ii)
(observing only the remainders)
3mn --Remainder--> (12*6)%18
=> mn --Remainder--> 24%18 = 6
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