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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]

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03 Aug 2006, 12:29
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If t is a positive integer and r is the remainder when t^2+5t+6 is divided by 7, 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

Can anyone tell me what the answer and how to arrive at the answer?
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03 Aug 2006, 13:13
(D).

ST1 => t=7k+6, for some integer k

Plugging into t^2+5t+6 we get

(7k+6)^2+5(7k+6)+6
49k^2+84k+36+35k+30+6
(49k^2+84k+35k)+72

Note, I grouped all members divisible by 7 in parethesis. So, the remainder must come from 72 when dividing by 7, which is 2. /SUFF/

ST2. => t^2=7k+1

Let's represent t=7p+j, then t^2=49i^2+14pj+j^2. We know that t^2 has a remainder 1 when it is divided by 7, therefore j^2=1, and j=1.

So, t=7p+1

Now, plugging back into the original equation we can again determine the remainder. So, ST2 also sufficient.

(D)
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03 Aug 2006, 13:31
Is it D??

We have t^2+5t+6........EQ1

St1:
t = 7x + 6 where x is a +ve integer. Putting this in eq1 we get

(7x+6)^2 + 5(7x+6) + 6
= 49x^2 + 84x+ 36 + 35x+ 30 + 6
= 49x^2 + 84x+35x+ 72
when this is divided by 7 then remainder will be 2 because terms in bold are fully divisible by 7: SUFF

St2:
t^2 = 7y+1 where y is a +ve integer. Let t = 7z+k

so t^2 = 49z^2 + k^2 + 14k = 7y+1
so k = 1
now we have t = 7z+1, Putting this in eq1 we get

7y+1 + 5*(7z+1) + 6
= 7y+1+35z+5+6 = 7y+35z + 12
terms in bold are fully divisible by 7. So remainder will be 5.: SUFF

But I have seen any GMAT or OG question that leads to different answers.
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03 Aug 2006, 18:53
(A)??? What source?

Anybody cares to explain/disprove why ps_dahya and my solutions for ST2 are wrong?
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03 Aug 2006, 19:01
Nevermind. I found my mistake.

ST2

Let's represent t=7p+j, then t^2=49i^2+14pj+j^2. We know that t^2 has a remainder 1 when it is divided by 7, therefore j^2=1, and j=1.

Piece in RED is wrong. It does not mean that j^2=1. It means that when j^2 is divided by 7 the remainder will be 1. Two different things!

j can be either 1 or 6. (1^2 and 6^2 both have remainder 1 when dividing by 7.) Hence, not sufficient!!!!

This one trapped me
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03 Aug 2006, 19:56
v1rok wrote:
Nevermind. I found my mistake.

ST2

Let's represent t=7p+j, then t^2=49i^2+14pj+j^2. We know that t^2 has a remainder 1 when it is divided by 7, therefore j^2=1, and j=1.

Piece in RED is wrong. It does not mean that j^2=1. It means that when j^2 is divided by 7 the remainder will be 1. Two different things!

j can be either 1 or 6. (1^2 and 6^2 both have remainder 1 when dividing by 7.) Hence, not sufficient!!!!

This one trapped me

You know I first answered A and then after seeing your post I changed mine to D. I handled statement 2 differently.
This was my initial analysis:
t^2 = 7y+1 where y is a +ve integer.

Now EQ1 becomes
49y^2 + 14y + 1 + 5 * SQRT(7y+1) + 6
= 49y^2 + 14y + 7 + 5 * SQRT(7y+1)
Bold part is fully divisible so drop that part. Now we have to take care of 5 * SQRT(7y+1)
7y+1 should be a perfect square
when y = 0 -------> 5 * SQRT(7y+1) = 5--------> Remainder = 5
when y = 5 -------> 5 * SQRT(7y+1) = 30--------> Remainder = 2: INSUFF
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03 Aug 2006, 20:07
statement A alone.

In statement B, the base t is a square hence we need more info in such cases. We get a quadratic equation and get two different values for the variable t and two different remainders consequently.
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03 Aug 2006, 23:19
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A

1) t = 7q+6

Lets say t = 6

Then t^2 + 5T + 6 = 72
remainder is 2

when t = 13
Then t^2 + 5T + 6 = 240
Remaninder = 2
Suff

2) t^2 = 7q + 1
hence t can be 6 or 8 etc

when t= 6 Remainder = 2
When t = 8 Remainder = 5

hence Not suff
03 Aug 2006, 23:19
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