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t^2 + 5t +6

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t^2 + 5t +6

by mberkowitz » Mon Oct 06, 2008 1:33 pm
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 is divided by 7 the remainder is 6.
2) when t^2 is divided by 7 the remainder is 1.

Ok.
If I saw this Q on the real thing I'd be sweating.

i know it was already spoken about some time ago, but can someone please elaborate?

so 1) I can say t=7a+6. plugging that into the equation from the stem for t (7a+6)^2 + 5 (7a+6) + 6 = 49a^2 + 56a + 63a + 72. i understand that all terms here are divisible by 7 EXCEPT 72, and 72/7 equals remander remainder 2. Therefore s1 gives us the answer.

As for statement 2, i suppose we could do the same sort of thing, which would take me a while, and find that statement 2 is insufficient. The answer is A.

is there a better method to solve it?

thanks very much, Mo.
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by Morgoth » Mon Oct 06, 2008 4:16 pm
First, solve the equation

t^2 + 5t + 6 = (t+2)(t+3)

statement (1)

t = 7q + 6
t can also be written as 7k-1, this means any multiple of 7 and then subtract 1 to the same when divided by 7 will always give a remainder 6

therefore,

(t+2)(t+3) = (7k-1+2)(7k-1+3) = (7k+1)(7k+2)

when (7k+1) will be divided by 7 the remainder will be 1
when (7k+2) will be divided by 7 the remainder will be 2

when the product of both is divided by 7 the remainder will be 2 for any value of k.
Sufficient.

Statement (2)
t^2 when divided by 7 the remainder is 1
We cannot follow the same strategy as above in this statement because here we are dealing with addition as compared to multiplication as statement (1)

this statement is even more simpler than the first one

t^2 + 5t + 6
t^2 = multiple of 7 + 1

t^2 + 6 = multiple of 7.

Therefore, we only have to find out if 5t/7 gives standard remainder across all cases.

5*5/7 = 35/7 = remainder 0
5*4/7 = 20/7 = remainder 6
5*6/7 = 30/7 = remainder 2.

Insufficient.

Thus, A is the answer.


Hope this helps.
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by eshahid » Mon Oct 06, 2008 4:18 pm
you are correct about '1'. Its gives us the asnwer.
so eliminate BCE and we are left with AD
consider '2'. t^2 divided by 7 gives remainder 1
that means 't' will be of the form 7a+1 or 7a-1
7a-1 can also be written as 7b+6
7a+1 and 7b+6 when substituted in '2' give different answers. So B is not sufficient.

Hope this helps
Shahid E

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by stop@800 » Mon Oct 06, 2008 9:58 pm
from A you can say t+1 is divisible by 7

t2+5t+6
t2+t+4t+6
t(t+1)+4t+4+2
t(t+1)+4(t+1)+2

as t+1 is divisible by 7
so remainder is 2
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by vivek.kapoor83 » Tue Oct 07, 2008 12:16 am
Supereb explanation @800..
another way could be (If you ahvent got brains like @800... :) )
by putting some values..
like for 1...u can take something that is divided by 7 leaves 6....like 13,20 and put it in eq...u ll see remainder is always 2.
b. you can take t =6 or8 and see the diff remainders..

So. A
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by mberkowitz » Tue Oct 07, 2008 6:13 am
thanks vivek, plugging numbers will do. reality is i dont have quant focus like many of you and badly need these "simple" strategies..
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by bluementor » Tue Oct 07, 2008 7:41 am
mbekorwitz, i'm going to piggy back on your post. hope you don't mind.

Suppose a more simplified question (assume this to be of PS type) was given:

For t>0, if t divided by 7 gives a remainder of 6, what is the remainder when t^2 is divided by 6?


How would you solve this?

BM
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by mberkowitz » Tue Oct 07, 2008 8:14 am
BM,

Im pretty sure, given the fact that t can equal 13 or 20, that the remainder of t^2 will never be a constant (as was proved in my query).

Therefor you can't see this question on the test. If, however, t^2 did have a constant remainder, I think solving this would hinge on your ability to understand that finding the remainder of t^2 lies in finding values for t, and then plugging them in.

Smart guys what do you all think?

mb
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by cubicle_bound_misfit » Tue Oct 07, 2008 12:09 pm
nicely done morgoth

Light from the art of sinking :P
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by maihuna » Sat May 09, 2009 11:04 am
mberkowitz wrote:BM,

Im pretty sure, given the fact that t can equal 13 or 20, that the remainder of t^2 will never be a constant (as was proved in my query).

Therefor you can't see this question on the test. If, however, t^2 did have a constant remainder, I think solving this would hinge on your ability to understand that finding the remainder of t^2 lies in finding values for t, and then plugging them in.

Smart guys what do you all think?

mb
It is an official question from GMAT FOCUS...
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