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If t is a positive integer and r is the remainder when

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If t is a positive integer and r is the remainder when [#permalink] New post 11 Jul 2009, 17:38
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If t is a positive integer and r is the remainder when t^2+5t+6 is divided by7, what is the value of r?

1. When t divided by 7, the remainder is 6

2. When t^2 divided by 7, the remainder is 1

OA after some discussion
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Re: Tough one [#permalink] New post 11 Jul 2009, 18:02
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stmt 1: if T gives r=6 when divided by 7, then T^2 will give r^2

but 6^2= 36 is greater than 7, so upon dividing 36 by 7, we get 1 as remainder.

so T^2 will finally give the remainder of 1

5T will give 5*6, again doing same as above, 30/7 will give 2 as remainder. So remainder from 5T is 2

6 will obviously give remainder of 6

adding all 3, (bcoz t^2, 5t and 6 have been added) we get 1+2+6 = 9 as remainder.

again divide 9 by 7 you get 2 as remainder. So you know the remainder if this info is given.

stmt 2 is insufficient.


NOTE: you dont need to solve for r, i showed the calculations so that one can understand how this has to be worked out.
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Re: Tough one [#permalink] New post 11 Jul 2009, 18:06
Just wanted to add that I didn't know any way how stmt 2 could give an answer, but just in case there could be a possibility, and since i had no logic to check, i took random numbers satisfying stmt 2, in this case 6 and 8, and substituted in the polynomial expression, both gave different remainders, so i was sure that stmt 2 has to be insufficient.
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Re: Tough one [#permalink] New post 12 Jul 2009, 07:22
sher676 wrote:
If t is a positive integer and r is the remainder when t^2+5t+6 is divided by7, what is the value of r?

1. When t divided by 7, the remainder is 6

2. When t^2 divided by 7, the remainder is 1

OA after some discussion


t^2+5t+6 = (t+3)(t+2) , devide by 7 (t+3)(t+2)/7 = (t/7 +3/7)(t/7+2/7)

from 1 and only considering remainders from the above the first bracket will always give 6+3 remaider and the 2nd 6+2

multiply together ( 9*8) = 72 , remainder is 2.........suff

from 2 ....insuff
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Re: Tough one [#permalink] New post 13 Jul 2009, 07:13
Guys, this is a very interesting question , I have seen this type for the first time. I have couple of questions

1) if remainder for t/7 is known we can know the remainder for t^2/7 , and vice verca is not possible ?
ie if the the remainder for t^2/7 is known t/7 is not known ?

2) where can I get additional information on these types of questions ?

Help !
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Re: Tough one [#permalink] New post 13 Jul 2009, 07:53
skpMatcha wrote:
Guys, this is a very interesting question , I have seen this type for the first time. I have couple of questions

1) if remainder for t/7 is known we can know the remainder for t^2/7 , and vice verca is not possible ?
ie if the the remainder for t^2/7 is known t/7 is not known ?

2) where can I get additional information on these types of questions ?

Help !


Actually there is a remainder rule, that works only for additions, subtractions and multiplications, but not for divisions.

Ill illustrate this rule with an example.

25/7 - R= 4
41/7 - R= 6
addition
(25+41)/7 should give remainder of 4+6, but 10 is greater than 7, so the final remainder will be that of 10/7, ie 3
checking this rule - 25+41 = 66..... 66/7 - R = 3

subtraction
(41-25)/7 should give remainder 6-4 = 2...... check: 41-25 = 16. .....16/7 - R= 2

multiplication
41*25/7 should give remainder of (6*4)/7, ie 24/7 -> R= 3
check: 41*25 = 1025....1025/7 -> R= 3


Now square is nothing but (t/7) * (t/7), so applying multiplication rule, we get that remainder should be R*R

Square root is division, and therefore the rule doesnt apply to square roots.

Hope this helps :)
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Re: Tough one [#permalink] New post 13 Jul 2009, 08:31
Thanks a bunch for the explanation. You put in a very simple way ,and I got it.

Quote:
Square root is division, and therefore the rule doesnt apply to square roots.


Is this the reason B is not sufficient ?
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Re: Tough one [#permalink] New post 13 Jul 2009, 08:33
skpMatcha wrote:
Thanks a bunch for the explanation. You put in a very simple way ,and I got it.

Quote:
Square root is division, and therefore the rule doesnt apply to square roots.


Is this the reason B is not sufficient ?


yes, but i wasn't sure so i also checked using random numbers.
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Re: Tough one [#permalink] New post 13 Jul 2009, 09:15
Quote:
If t is a positive integer and r is the remainder when t^2+5t+6 is divided by7, what is the value of r?

1. When t divided by 7, the remainder is 6

2. When t^2 divided by 7, the remainder is 1


2. When t^2 divided by 7, the remainder is 1

The remainder is 1

It is possible only when the remainder is 1 or 6 (when t/7)

when remainder is 1
(7+1)*(7+1) =8x8 =64/7--> remainder is 1
(14+1)*(14+1)=15x12 =225/7--> remainder is 1

when remainder is 6
(7+6)*(7+6)=13x13-->169/7 -->remainder is 1

Therefore not sufficient...
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Re: Tough one   [#permalink] 13 Jul 2009, 09:15
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