If t is a positive integer and r is the remainder when t^2 + 5t + 6 is

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 (\(t^2+5t+6=(t+2)(t+3)=(7q+8)(7q+9)\)).

Hope it's clear.

My approach:

From (1) - t/7 gives remainder = 6. So we can easily find remainder for t^2 and 5*t in the algebraic exp. t^2 + 5t + 6 using properties of remainder.

Therefore the remainder of algebraic exp. = 1 + 2 + 6 = 9 (7 + 2) = 2 ................. [A]


Properties of remainder:

[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.

Thus A is the answer.

(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

FROM STATEMENT - I INSUFFICIENT

The minimum value of t = 13

So, \(t^2+5t+6 = 13^2+5*13+6\) = \(240\)

Now, \(\frac{240}{7}\) = Remainder 2

FROM STATEMENT - II INSUFFICIENT

The minimum value of t = +1 and -1

So, \(t^2+5t+6 = 1^2+5*1+6\) = \(12\) Remainder 5

And \(t^2+5t+6 = -1^2+5*-1+6\) = \(2\) Remainder 2

Thus, Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, hence correct answer will be (A)

Hi Bunuel ,

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.

Bunuel For st1) Why can we not replace (7p+6) for all values of t?

You CAN replace t with 7p + 6 in t^2 + 5t + 6. You'll get 7*something + 72. 7*something is divisible by 7, and 72 yields the remainder of 2 upon division by 7.

The patterns in remainder problems will always emerge fairly early when you plug in numbers. therefore, if you don't IMMEDIATELY realize a good theoretical way to do a remainder problem, you should get on the number plugging RIGHT AWAY.


I would IMMEDIATELY start plugging in numbers and using PATTERN RECOGNITION on a problem like this one. remainder problems usually show patterns after a very, very small number of plug-ins.

Statement (1):
it's easy to generate t's that do this: 6, 13, 20, 27,

... (note that 6 is a member of this list, and an awfully valuable one at that; it's quite easy to plug in)
try 6: 36 + 30 + 6 = 72; divide by 7 --> remainder 2
try 13: 169 + 65 + 6 = 240; divide by 7 --> remainder 2
try 20: 400 + 100 + 6 = 506; divide by 7 --> remainder 2
by this point i'd be convinced.

Note that 3 plug-ins is NOT good enough for a great many problems, esp. number properties problems.

however, as i said above, remainder problems don't keep secrets for long.
sufficient.



Statement (2):
it's harder to find t's that do this.
however, the gmat is nice to you. if examples are harder to find, then the results will usually come VERY quickly once you find those examples.
just take perfect squares, examine them, and see whether they give the requisite remainder upon division by 7.
the first two perfect squares that do so are 1^2 = 1 and 6^2 = 36.
if you don't recognize that (1^2)/7 gives remainder 1, then you'll have to dig up 6^2 = 36 and 8^2 = 64. that's not that much more work.
in any case, you'll have:

For 1 & 6:
1 + 5 + 6 = 12 --> divide by 7; remainder = 5
36 + 30 + 6 = 72 --> divide by 7, remainder = 2 (the work for this was already done above; you should NOT do it twice. i'm reproducing it here only for the sake of quick understanding.)

or

For 6 & 8:
36 + 30 + 6 = 72 --> divide by 7, remainder = 2 (the work for this was already done above; you should NOT do it twice. i'm reproducing it here only for the sake of quick understanding.)
64 +40 + 6 = 110 --> divide by 7, remainder = 5 either way, insufficient within the first two plug-ins!

answer (a)

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Bunuel Hi,

Can you pls explain why in St. 1 if you solve it alternatively and take t to be 6,13,20,etc. (using the n = r, r +d, r
+2d, etc.. formula (where r is the remainder and d the divisor ) and replace each of those values in :

the equation (t+3)(t+2) / 7 you get different remainders then st. is not sufficient right?

Pls help understand why this method is wrong? Thanks.

If t = 6, then t^2 + 5t + 6 = 72. 72 divided by 7 gives the remainder of 2.
If t = 13, then t^2 + 5t + 6 = 240. 240 divided by 7 gives the remainder of 2.
If t = 20, then t^2 + 5t + 6 = 506. 506 divided by 7 gives the remainder of 2.
...

Hi Bunuel,

re the stat 2, may I ask is possible to use the property of remainders that says the remainder of a product is the product of the remainders of individual factor fraction.
And since rem ((t^2)/7) is 1, it follows that rem (t/7) = 1, so rem ((5*t)/7) is equal to rem (5/7)*rem (t/7), i.e. 5*1 so 5. Then rem (6/7) = 6. And the rem of the entire fraction is 1+5+6 = 12?

First of all, the rule you are citing is not entirely accurate:

Let N1, N2, N3,... be positive integers that, when divided by a positive integer D, produce remainders R1, R2, R3,... respectively. The remainder of the product of N1*N2*N3*... when divided by D is the same as the remainder when the product of R1*R2*R3*... is divided by D.

For example, say N1 = 5, N2 = 7, and N3 = 10

Let's take the positive integer divisor D = 4.

• When N1 = 5 is divided by D = 4, the remainder R1 is 1.
• When N2 = 7 is divided by D = 4, the remainder R2 is 3.
• When N3 = 10 is divided by D = 4, the remainder R3 is 2.

Now, consider the product of these numbers:

• N1 * N2 * N3 = 5 * 7 * 10 = 350

When 350 is divided by 4, the remainder is 2.

On the other hand:

• R1 * R2 * R3 = 1 * 3 * 2 = 6

When 6 is divided by 4, the remainder is also 2.

However, this property doesn't work in reverse. That is, knowing the remainder of the product doesn't always let you determine the remainders of the individual factors. Hence, the fact that t*t divided by 7 gives a remainder of 1 doesn't necessarily mean that t divided by 7 will also give a remainder of 1. For example, consider t = 1 and t = 6. t^2 in both cases gives a remainder of 1 when divided by 7; however, t divided by 7 gives a remainder of 1 in one case and a remainder of 6 in another.

Hope it's clear.

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