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.
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.
Can anyone shed some light on this type of problem, specifically a better method to solve it?
thanks very much, Mo.
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- Guest
- Guest
This is a pretty simple problem. Just look at it calmly.
From 1:
when t is divided by 7 the remainder is 6.
Now look at the stem:
t^2+5t+6 is divided by7 = (t^2 / 7) +(5t /7)+6/7
Now if t/7 gives us 6,
t^2/7 will give us 1 ..... always (try it)
and we know what 6/7 will lead to.
From 2:
Although it is clearly stated t^2/7 leaves 6, the same cannot be said for t.....(sign of t will affect the result, as it is to be added to two other numbers)
From 1:
when t is divided by 7 the remainder is 6.
Now look at the stem:
t^2+5t+6 is divided by7 = (t^2 / 7) +(5t /7)+6/7
Now if t/7 gives us 6,
t^2/7 will give us 1 ..... always (try it)
and we know what 6/7 will lead to.
From 2:
Although it is clearly stated t^2/7 leaves 6, the same cannot be said for t.....(sign of t will affect the result, as it is to be added to two other numbers)
- mbarshaik
patience
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.
From the question statement.
t^2+5t+6 = 7k + r
(t + 3) (t + 2) = 7k + r
Now from 1 -
t = 7n + 6; Lets substitute this in the above equatio and it gives
(7n + 9) (7n + 8) = 7k + r
So i know that above numbers are consecutive, so lets substitute n =1
16 * 15 = 7k + r
Divide 16 * 15 by 7 you will be left with 2.
Try putting n = 2 or some other number you will reach to same conclusion .
Now from Statement 2 -
when t^2 is divided by 7 the remainder is 1
t^2 = 7n + 1
Put n = 1, t^2 = 9 then t = +3 or -3
Lets put it in main equation (t + 3) (t + 2) = 7k + r
we get two different answers. 0 & 2 as remainders.
So from this we can conclude A is teh answer.
1) when t is divided by 7 the remainder is 6.
2) when t^2 is divided by 7 the remainder is 1.
From the question statement.
t^2+5t+6 = 7k + r
(t + 3) (t + 2) = 7k + r
Now from 1 -
t = 7n + 6; Lets substitute this in the above equatio and it gives
(7n + 9) (7n + 8) = 7k + r
So i know that above numbers are consecutive, so lets substitute n =1
16 * 15 = 7k + r
Divide 16 * 15 by 7 you will be left with 2.
Try putting n = 2 or some other number you will reach to same conclusion .
Now from Statement 2 -
when t^2 is divided by 7 the remainder is 1
t^2 = 7n + 1
Put n = 1, t^2 = 9 then t = +3 or -3
Lets put it in main equation (t + 3) (t + 2) = 7k + r
we get two different answers. 0 & 2 as remainders.
So from this we can conclude A is teh answer.
- MBARSHAIK
Patience
Oops...in the solution for statement 2 i messed up with addition. It cant be 9, it shud be 8. even tehn the solution holds good.
- vik
1. You can pick any positive number for t which when divided by 7 has 6 as a reminder. (Eg. 6,20)
It always leaves a reminder of 2 for the equation t^2+5t+6. Hence sufficient.
2. Here when you pick different numbers the reminder r for the entire equation chenges.
For eg. when t = 1, 1^2 + 5*1 + 6 = 12 which when divided by 7 leaves 5 as the reminder.
Whereas when t = 6, 6^2 + 5*6 + 6 = 72 which when divided by 7 leaves 2 as the reminder.
Hence insufficient.
It always leaves a reminder of 2 for the equation t^2+5t+6. Hence sufficient.
2. Here when you pick different numbers the reminder r for the entire equation chenges.
For eg. when t = 1, 1^2 + 5*1 + 6 = 12 which when divided by 7 leaves 5 as the reminder.
Whereas when t = 6, 6^2 + 5*6 + 6 = 72 which when divided by 7 leaves 2 as the reminder.
Hence insufficient.
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
ok dudes. i scored pretty high on this thing, and 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 ÷ 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
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
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)
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 ÷ 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
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
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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