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When positive integer n is divided by 3, the remainder is 2;

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When positive integer n is divided by 3, the remainder is 2; [#permalink]

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27 Aug 2006, 02:18
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When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

1) n-2 is divisible by 5

2) t is divisible by 3

OPEN DISCUSSION OF THIS QUESTION IS HERE: when-positive-integer-n-is-divided-by-3-the-remainder-is-86155.html
[Reveal] Spoiler: OA

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27 Aug 2006, 03:36
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[quote="yezz"]When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

1) n-2 is divisible by 5

2) t is divisible by 3

How can we generalize this concept ????

Thanks[/quote]
I think C it is
(1) alone
we have n-2 is divisible by both 3 and 5 so n=15a+2
t=5b+3 so nt=(15a+2)*(5b+3)=75ab+45a+10b+6
This tells nothing
(2) alone
t=15c+3 so nt=(3d+2)*(15c+3)=45ab+30b+9a+6
This tells nothing also
(1)(2) together we have
nt= (15a+2)*(15c+3)=15N+6
This means the remainder when nt is divided by 15 is 6
C it is

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27 Aug 2006, 03:43
yezz wrote:
When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

1) n-2 is divisible by 5

2) t is divisible by 3

How can we generalize this concept ????

Thanks

From question, we know that n-2 is a multiple of 3 and t-3 is a multiple of 5.

(1) says that n-2 is a multiple of 5. We know that n-2 is a multiple of 3, so n-2 is a multiple of 15 and n is 2 higher than a multiple of 15. Thus when n is divided by 15, the remainder is 2. All we know about t is that its units digit is either 3 or 8. If t=3, the remainder when nk is divided by 15 will be 6. If t=8, the remainder when nk is divided by 15 will be 1. NOT SUFF

(2) t is a mulltiple of 3. If n is a multiple of 5, nk will be a multiple of 15. If n is not a multiple of 5, mk will not be a multiple of 15. So when nk is divided by 15, the remainder may or may not be 0 NOT SUFF Note that As t is 3 more than a multiple of 5, the remainder when t is divided by 15 is either 3,8,or 13. But t is a multiple of 3, so when t is divided by 15, the remainder is 3,6,9 or 12. Thus from (2), we know that t/15 yields a remainder of 3.

Combining (1) and (2) n=15x+2 and t=15y+3 for some non-negative integers x and y. nt= 15^2xy+3(15x)+2(15y)+6. Each of the first three terms is a multiple of 15, so 6 is the remainder when nt is divided by 15

SUFF

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27 Aug 2006, 03:52
What if we assumed hypothetically that we have a number x say
a) when it is devided by 3 remainder is 2
b) when divided by 5 remainder is 1

This number is sure devisible by 15 but how can we know the remainder in this case when devided by 15.

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27 Aug 2006, 11:53
Kevin , I do understand why n when devided by 15 the remainder will always be 2.

But what if in the question stem , n

a) when it is devided by 3 remainder is 2
b) when divided by 5 remainder is 1

how can we know the remainder here?

and what if x is an intiger devisible by 4 and when divided by 5 yields a remainder of 3 ( is there a general formula for this number if we want to know the remainder when we divide it by 20 ie ( 5*4)
am really confused wn i try to generalize the way you attacked the problem???

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27 Aug 2006, 15:31
yezz wrote:
Kevin , I do understand why n when devided by 15 the remainder will always be 2.

But what if in the question stem , n

a) when it is devided by 3 remainder is 2
b) when divided by 5 remainder is 1

how can we know the remainder here?

and what if x is an intiger devisible by 4 and when divided by 5 yields a remainder of 3 ( is there a general formula for this number if we want to know the remainder when we divide it by 20 ie ( 5*4)
am really confused wn i try to generalize the way you attacked the problem???

Think that between two multiples of 15, there are 4 multiples of 3
15k, 15k+3,15k+6,15k+9,15k+12, 15(k+1)

So if n yields a remainder of 2 when divided by 3, when divided by 15, n could yield a remainder of 5,8,11 or 14

Similarly, if when divided by 5, n yields a remainder of 1, when divided by 15, n will yield a remainder of 1,6,or 11

If both of these are true, n must yield a remainder of 11 when divided by 15.

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27 Aug 2006, 15:39
Quote:
and what if x is an intiger devisible by 4 and when divided by 5 yields a remainder of 3 ( is there a general formula for this number if we want to know the remainder when we divide it by 20 ie ( 5*4)

Since 4 is a factor of 20, if x is divisible by 4, x yields a remainder of 4k when divisible by 4 (here k is an integer from 0 to 4)

Since 5 is also a factor of 20, and when x is divided by 5, the remainder is 3, when x is divided by 20, the remainder is 5m+3, where m is from 0 to 3

So can 4k=5m+3 if k belongs to {0,1,2,3,4} and m belongs to {0,1,2,3}?

k=(5m+3)/4, so 5m+3 must be a multiple of 4- m must be 1 and the remainder when x is divided by 20 is 5(1)+3=8

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27 Aug 2006, 22:31
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This method may help.

Let n = 3x+2........EQ1
t = 5y+3..............EQ2
nt = 15xy + 10y + 9x+6
Remove 15xy because this is fully divisible by 15.
Now we have to find remainder when 10y + 9x+6 (.........EQ3) is divided by 15.

St1: n-2 is divisible by 5. From EQ1 we get n-2 = 3x for 3x to be divisible by 5 x must be a multiple of 5.
So EQ3 reduces to 10y+9*5z + 6
9*5z is divisible by 15. So removing this we get 10y+6. No conclusion: INSUFF

St2: t is divisible by 3. From EQ2, if t is to be divisible by 3 then y must be a multiple of 3. So EQ3 reduces to 10*3w+9x+6. 10*3w is divisible by 15. Removing this we get 10y+6. No conclusion: INSUFF

Together:
EQ3 reduces to 10*3w + 9*5z + 6. Both 10*3w and 9*5z are divisible by 15. Hence 6 will be remainder: SUFF
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28 Aug 2006, 02:39
Folks ( Dahiya and Kevin )you are great.... it takes me some time to grasp new concepts but you really helped a lot .

I merely spent the whole day with this issue in mind trying to blug in different combinations of numbers to find the concept ...but without your help i could ve never make it .

Thanks a lot

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Re: When positive integer n is divided by 3, the remainder is 2; [#permalink]

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09 Nov 2013, 07:51
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Re: When positive integer n is divided by 3, the remainder is 2;   [#permalink] 09 Nov 2013, 07:51
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