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

[Raj]

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

Answer is A.

From (1), I could figure (t+3)(t+2) will always have a remainder 2, hence SUFFICIENT

I had trouble with (2) as I could not come up with an algebraic approach. I understand I can plug numbers to see that t^2 = 36 and t^2 = 64 fit the criterion BUT yield different remainders when the corresponding values of t are plugged into t^2 + 5t +6 and hence INSUFFICIENT.

Is there a more elegant, less time consuming approach for this problem?
-Raj.

[Mink]

7 is a prime (no factors) ; so although we know about : t^2 in the equation t^2+5t+6; we can really not figure out anything about t.

Given t you can figure out t^2 but not vice versa in case of a prime.
Hope this helps.

[Guest]

Does someone mind explaining this in a bit more detail?

[RonPurewal]

Does someone mind explaining this in a bit more detail?

one fact that's pretty cool, and which happens to apply to this problem, is that you can do normal arithmetic with remainders, as long as all the remainders come from division by the same number. the only difference is that, if/when you get numbers that are too big to be authentic remainders (i.e., they're equal to or greater than the number you're dividing by), you have to take out as many multiples of the divisor as necessary to convert them back into "legitimate" remainders again. you can think of the remainders as on an odometer that rolls back to 0 whenever you reach the number you're dividing by.

so with statement (1), all the remainders are upon division by 7, so we can do normal arithmetic with them:
if t gives a remainder of 6, then t^2 = t x t gives a remainder of 6 x 6 = 36 --> this is more than 7, so we take out as many 7's as possible: 36 - 35 = 1.
if t gives a remainder of 6, then 5t gives a remainder of 5(6) = 30 --> this is more than 7, so we take out as many 7's as possible: 30 - 28 = 2.
and finally, 6 itself gives a remainder of 6.
therefore, the grand remainder when t^2 + 5t + 6 is divided by 7 should be 1 + 2 + 6 = 9 --> take out one more seven --> remainder will be 2.
sufficient.

by the way, much more generally (and therefore perhaps more importantly), 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.

with statement (1), generate the first 3 numbers for which the statement is true: 6, 13, 20.
try 6: 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
try 13: 169 + 65 + 6 = 240, which yields a remainder of 2 upon division by 7.
try 20: 400 + 100 + 6 = 506, which yields a remainder of 2 upon division by 7.
i'm convinced. (again, remember that PATTERNS EMERGE EARLY in remainder problems. 3 examples may not be enough for other types of pattern recognition, but that's usually pretty good in a remainder problem.)

with statement (2), as a poster has already mentioned above, find the first two t^2's that actually do this, which are 1^2 = 1 and 6^2 = 36.
if t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.
if t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
insufficient.

if you know about 'modulo arithmetic' (see this post), then you can shortcut this sort of problem with it.

[victorgsiu]

the modulo concept is excellent - this makes plugging in much easier, as you can now divide each component of t^2+5t+6 by 7, add the remainders, and divide again by 7 to get the "Master Remainder".

Does someone mind explaining this in a bit more detail?

one fact that's pretty cool, and which happens to apply to this problem, is that you can do normal arithmetic with remainders, as long as all the remainders come from division by the same number. the only difference is that, if/when you get numbers that are too big to be authentic remainders (i.e., they're equal to or greater than the number you're dividing by), you have to take out as many multiples of the divisor as necessary to convert them back into "legitimate" remainders again. you can think of the remainders as on an odometer that rolls back to 0 whenever you reach the number you're dividing by.

so with statement (1), all the remainders are upon division by 7, so we can do normal arithmetic with them:
if t gives a remainder of 6, then t^2 = t x t gives a remainder of 6 x 6 = 36 --> this is more than 7, so we take out as many 7's as possible: 36 - 35 = 1.
if t gives a remainder of 6, then 5t gives a remainder of 5(6) = 30 --> this is more than 7, so we take out as many 7's as possible: 30 - 28 = 2.
and finally, 6 itself gives a remainder of 6.
therefore, the grand remainder when t^2 + 5t + 6 is divided by 7 should be 1 + 2 + 6 = 9 --> take out one more seven --> remainder will be 2.
sufficient.

by the way, much more generally (and therefore perhaps more importantly), 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.

with statement (1), generate the first 3 numbers for which the statement is true: 6, 13, 20.
try 6: 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
try 13: 169 + 65 + 6 = 240, which yields a remainder of 2 upon division by 7.
try 20: 400 + 100 + 6 = 506, which yields a remainder of 2 upon division by 7.
i'm convinced. (again, remember that PATTERNS EMERGE EARLY in remainder problems. 3 examples may not be enough for other types of pattern recognition, but that's usually pretty good in a remainder problem.)

with statement (2), as a poster has already mentioned above, find the first two t^2's that actually do this, which are 1^2 = 1 and 6^2 = 36.
if t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.
if t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
insufficient.

if you know about 'modulo arithmetic' (see this post), then you can shortcut this sort of problem with it.

[Ben Ku]

Hope your questions are answered.

[anjali.gmat]

Does someone mind explaining this in a bit more detail?

one fact that's pretty cool, and which happens to apply to this problem, is that you can do normal arithmetic with remainders, as long as all the remainders come from division by the same number. the only difference is that, if/when you get numbers that are too big to be authentic remainders (i.e., they're equal to or greater than the number you're dividing by), you have to take out as many multiples of the divisor as necessary to convert them back into "legitimate" remainders again. you can think of the remainders as on an odometer that rolls back to 0 whenever you reach the number you're dividing by.

so with statement (1), all the remainders are upon division by 7, so we can do normal arithmetic with them:
if t gives a remainder of 6, then t^2 = t x t gives a remainder of 6 x 6 = 36 --> this is more than 7, so we take out as many 7's as possible: 36 - 35 = 1.
if t gives a remainder of 6, then 5t gives a remainder of 5(6) = 30 --> this is more than 7, so we take out as many 7's as possible: 30 - 28 = 2.
and finally, 6 itself gives a remainder of 6.
therefore, the grand remainder when t^2 + 5t + 6 is divided by 7 should be 1 + 2 + 6 = 9 --> take out one more seven --> remainder will be 2.
sufficient.

by the way, much more generally (and therefore perhaps more importantly), 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.

with statement (1), generate the first 3 numbers for which the statement is true: 6, 13, 20.
try 6: 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
try 13: 169 + 65 + 6 = 240, which yields a remainder of 2 upon division by 7.
try 20: 400 + 100 + 6 = 506, which yields a remainder of 2 upon division by 7.
i'm convinced. (again, remember that PATTERNS EMERGE EARLY in remainder problems. 3 examples may not be enough for other types of pattern recognition, but that's usually pretty good in a remainder problem.)

with statement (2), as a poster has already mentioned above, find the first two t^2's that actually do this, which are 1^2 = 1 and 6^2 = 36.
if t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.
if t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
insufficient.

if you know about 'modulo arithmetic' (see this post), then you can shortcut this sort of problem with it.

Hi Ron,

I thought divisibility rules work for multiplication as well. So if t^2 gives a remainder of 1. I assumed t would give a remainder of 1. 1 * 1 = 1. Why does the rule not work here? Thanks.

[RonPurewal]

I thought divisibility rules work for multiplication as well. So if t^2 gives a remainder of 1. I assumed t would give a remainder of 1. 1 * 1 = 1. Why does the rule not work here? Thanks.

it doesn't work because a remainder is not the original number. i.e., in your example, 1 is a single number, but there are infinitely many numbers that will give remainders of 1 upon division.

for instance (considering division by 10 instead of division by 7), the remainder of 19/10 is 9, but the remainder of 19^2/10 =361/10 is 1.