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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?

First of all factor \(t^2+5t+6\) --> \(t^2+5t+6=(t+2)(t+3)\).

(1) When t is divided by 7, the remainder is 6 --> \(t=7q+6\) --> \((t+2)(t+3)=(7q+8)(7q+9)\). Now, no need to expand and multiply all the terms, just notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient.

(2) When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 or t=6, which when substituted in \((t+2)(t+3)\) will give different remainder upon division by 7. Not sufficient.

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?

First of all factor t^2+5t+6 --> t^2+5t+6=(t+2)(t+3)

(1) When t is divided by 7, the remainder is 6 --> t=7q+6 --> (t+2)(t+3)=(7q+8)(7q+9). Now, no need to expand and multiply all the terms, just notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient.

(2) When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 or t=6, which when substituted in (t+2)(t+3) will yield different remainder upon division by 7. Not sufficient.

Answer: A.

Hope it's clear.

Sorry Bunuel but this part highlighted is not completely clear: on the left side of = we have the two roots of our equation; on the right (t+2) ----> is 7q+6+2 ---> 7q+8 (same reasoning for 7q+9) BUT why we equal this two part ?? Can you explain me please ??

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?

First of all factor t^2+5t+6 --> t^2+5t+6=(t+2)(t+3)

(1) When t is divided by 7, the remainder is 6 --> t=7q+6 --> (t+2)(t+3)=(7q+8)(7q+9). Now, no need to expand and multiply all the terms, just notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient.

(2) When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 or t=6, which when substituted in (t+2)(t+3) will yield different remainder upon division by 7. Not sufficient.

Answer: A.

Hope it's clear.

Sorry Bunuel but this part highlighted is not completely clear: on the left side of = we have the two roots of our equation; on the right (t+2) ----> is 7q+6+2 ---> 7q+8 (same reasoning for 7q+9) BUT why we equal this two part ?? Can you explain me please ??

Thank you

We are asked to find the remainder when \(t^2+5t+6\) is divided by 7 or as \(t^2+5t+6=(t+2)(t+3)\), the remainder when \((t+2)(t+3)\) is divided by 7.

Now, from (1) we have that \(t=7q+6\). Substitute \(t\) with \(7q+6\) in \((t+2)(t+3)\) to get \((7q+8)(7q+9)\). So, finally we have that we need to find the remainder when \((7q+8)(7q+9)\) is divided by 7 (\(t^2+5t+6=(t+2)(t+3)=(7q+8)(7q+9)\)).

[1] If a quotient is squared, the remainder also gets squared and we need to adjust the new remainder based on no. we get on squaring.

Here quotient = t, r = 6. So, t^2 will have r' = 36 (7*5 + 1). Therefore r' =1

[2] If a quotient is multiplied by an integer k , the remainder also gets multiplied and we need to adjust the new remainder based on no. we get on multiplication.

Here quotient = t, r = 6. So, 5t will have r'' = 30 (7*4 + 2). Therefore r'' = 2

We get [A] using [1] and [2].

Clearly Stmt (2) can have multiple possibilities i.e. if t= 1 (t^2 =1) and t =6 (t^2 = 36) both give remainder 1, so no unique answer and hence rejected.

If t is a positive integer and r is the remainder when t^2+5 [#permalink]

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25 Aug 2014, 11:51

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(2) When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 or t=6, which when substituted in (t+2)(t+3) will give different remainder upon division by 7. Not sufficient.

Bunuel could you please explain why statement 2 is not sufficient and how the values for 't' could be either 1 or 6?

Re: If t is a positive integer and r is the remainder when t^2+5 [#permalink]

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06 Sep 2014, 04:17

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Shehryar Khan wrote:

(2) When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 or t=6, which when substituted in (t+2)(t+3) will give different remainder upon division by 7. Not sufficient.

Bunuel could you please explain why statement 2 is not sufficient and how the values for 't' could be either 1 or 6?

Hii Shehryar khan, let me try to explain..

t^2/7=1----------given

as u can see,we have 2 values of 2..1^2/7=remainder 1 and 6^2/7=36/7=remainder 1.. when we substitute 1 in (t+2)(t+3)/7 ,we get remainder as 5.. when we substitute 6 in (t+2)(t+3)/7, we get remainder as 2..

So,since the results are inconsistent,we cant have one definite remainder,which we are required to find,as per the stem..so,not sufficient..

Please consider KUDOS if my post helped
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If t is a positive integer and r is the remainder when t^2+5 [#permalink]

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04 Jan 2017, 03:14

I am sure that I am missing something because stem (1) tells that t=7q+6 so t=6,13,20,... then we can try different values of t so that we can find out the reminder, so it follows that (t+2)(t+3)=5p+r for t=13 (13+2)(13+3)=240, 240/5 gives a reminder of 0 for t=20 (20+2)(20+3)=506, 506/5 gives a reminder of 1. So stem (1) Not sufficient... Please, can someone explain to me what I am missing? Thanks

Re: If t is a positive integer and r is the remainder when t^2+5 [#permalink]

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11 Feb 2017, 22:58

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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.

As t^2 = 7m+1 so t^2 could be 1,8,15,22. Which gives the remainder as 5. But you have taken the value as 1 and 6. Could you please explain why you have taken 6 instead of 8 in case 2.

When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 ort=6, which when substituted in (t+2)(t+3)(t+2)(t+3) will give different remainder upon division by 7.
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As t^2 = 7m+1 so t^2 could be 1,8,15,22. Which gives the remainder as 5. But you have taken the value as 1 and 6. Could you please explain why you have taken 6 instead of 8 in case 2.

When t^2 is divided by 7, the remainder is 1 --> different values of t possible: for example t=1 ort=6, which when substituted in (t+2)(t+3)(t+2)(t+3) will give different remainder upon division by 7.

(2) reads: t^2 is divided by 7, the remainder is 1, not t.

t = 1 --> t^2 = 1 --> 1 divided by 7 gives the remainder of 1. t = 6 --> t^2 = 36 --> 36 divided by 7 gives the remainder of 1.
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