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Challenge - divisibility and exponents (m05q33) [#permalink]
 08 Nov 2007, 22:33
Question Stats:
86% (01:52) correct
13% (01:37) wrong based on 3 sessions
Is 4t^3 - 2t^2 - 8t + 16 divisible by t^2 ? 1. t \gt 12. t is an even prime number Source: GMAT Club Tests - hardest GMAT questions How can i solve this quickly and intuitively? I can certainly bang out the roots but i want to know the shortcut
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( 4t^3 - 2t^2 - 8t + 16 ) /t^2 = 4t-2- (8/t) +(16/t^2)
Thus you have to check whether 8/t and 16/t^2 is possible or not.
From (i)
t can be 2 and also 3. So may or may not be possible.
Insufficient
From (ii)
The only even prime number is 2. Which clearly divides the given polynomial.
Thus the answer is "B".
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
25 Oct 2012, 05:23
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bmwhype2 wrote: Is 4t^3 - 2t^2 - 8t + 16 divisible by t^2 ? 1. t \gt 12. t is an even prime number Source: GMAT Club Tests - hardest GMAT questions How can i solve this quickly and intuitively? I can certainly bang out the roots but i want to know the shortcut Since first two terms of 32t^3 - 16t^2 + 8t - 4 are divisible by t^2 then the question becomes whether 8t-4 is divisible by t^2. (1) t^2 <25 --> since t is a prime number then t=2 or t=3. If t=2 then 8t-4 is divisible by t^2=4 but if t=3 then 8t-4 is NOT divisible by t^2=9. Not sufficient. (2) t^2-8t+12=0 --> t=2 ( t=6 is not a valid solution since 6 is not a prime number). Sufficient. Answer: B.
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Re: Challenge - divisibility and exponents [#permalink]
08 Nov 2007, 22:51
bmwhype2 wrote: Is 4t^3 - 2t^2 - 8t + 16 divisible by t^2?
1. t > 1
2. t is an even prime
How can i solve this quickly and intuitively? I can certainly bang out the roots but i want to know the shortcut
Plugging in is probably the fastest method.
Start w/ Stat 2: t = 2, plug it in and you know it works. Answer is yes.
Stat 1: We already know that 2 works. Try 3 now. It doesn't work. Insuff.
Answer B
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my way to solve this would be..
1) basically tells us nothing..is t an integer or a fraction??
2) sufficient
the entire stem is divisible by 2(the only even prime)
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divide 4t^3-2t^2-8t+16 with t^2 and the fractions i have got are 8/t and 16/t^2. If,
A) t>1; t=2 or 3; Could be divisible or Could not be divisible...INSUFF
B) t is even prime so, t=2. Must be divisible....SUFF
ANS: B
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
22 Oct 2010, 06:04
The answer is B, as 2 is the only even prime number, and condition one gives us no real picture of the solution.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
22 Oct 2010, 12:16
1 tells us nothing. t>1 could mean anything; "1 + millionth part of 1" or "infinity" or "2". cant be sure. 2 tells us that t=2, whether or not 4 will divide the equation, we sure can find out. B it is.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
22 Oct 2010, 20:49
Since we need to divide the eqn with t^2,
t^2(4t-2-8/t+16/t^2)
When dividing with t^2, we get
(4t-2-8/t+16/t^2).
for this eqn to be int., 8/t and 16/t^2 must be int. only two values of t complies, 2 and 4.
Option1. no can be anything 2,3,4,5.. Insufficient
Option2. Only even prime no is 2. ...SUfficient.
Ans B.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
24 Oct 2010, 10:49
Let's start with statement B. Even prime is 2. so the statement B is sufficient.
statement 1: solve by plugging numbers. We know the given equation is divisible by 2. now lets take t=3 which give that the equation is not divisible.
So my answer is B.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
11 Nov 2010, 10:54
punzo wrote: Since we need to divide the eqn with t^2,
t^2(4t-2-8/t+16/t^2)
When dividing with t^2, we get
(4t-2-8/t+16/t^2).
for this eqn to be int., 8/t and 16/t^2 must be int. only two values of t complies, 2 and 4.
Option1. no can be anything 2,3,4,5.. Insufficient
Option2. Only even prime no is 2. ...SUfficient.
Ans B. This is a good way of tackling the question... I wasted a lot of time in trying to factorize the equation...
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
16 Nov 2010, 06:48
Isn't -2 also an even integer? Hence donn't we require both statements?
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
16 Nov 2010, 06:55
i guess prime numbers are only +ve
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
16 Nov 2010, 07:14
Thank you for your quick response. I am just starting out on my GMAT quest, so I appologise in advance for the potentially stupid question: wasn't the initial question asking whether the expression is divisible by t2. You have yourself have eloquently identified the requirements for the expression to be divisible by t2, however the information provided does not indicate at any point that t is in fact 2. In other words we require further information to narrow down t's value to be able to conclusicely say whether or not it is divisible. Posted from my mobile device 
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
16 Nov 2010, 07:18
t is an even prime number thus only poss val of t is 2
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
25 Oct 2011, 05:54
bmwhype2 wrote: Is 4t^3 - 2t^2 - 8t + 16 divisible by t^2 ? 1. t \gt 12. t is an even prime number Source: GMAT Club Tests - hardest GMAT questions How can i solve this quickly and intuitively? I can certainly bang out the roots but i want to know the shortcut stmt 1 gives us no suffş info. so A ,C and D are out stmt 2 - 2 is the only one even prime number. plugging in this number to 4t^3 - 2t^2 - 8t + 16, we can get an integer. so B is the answer
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
25 Oct 2011, 11:50
2) t=2 simple
1) pick t=10, then this 4t^3 - 2t^2 - 8t + 16 will be smth like 4000-200-80+16; don't compute. we simple need to have this divisible by t^2=100... therefore, since we have units and tens digits other than 0, it's not divisible by 100.
for t=2 suff; for t=10 not suff. => 1) is NOT suff.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
13 Nov 2011, 19:02
Start with B. The only even prime integer is 2. Plug in 2 for all T. Comes out to 24 and it is divisible by t^2 (4). For A, you already have 2 as a "yes." now, pick 3 and plug in and it is not divisible, hence "no". Answer is B.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]
25 Oct 2012, 06:16
I did it in this way
with little manipulations one gets 2t(2t^2 - t - 4) + 16...
from this we know that the number in question is even (....2t (2t^2 - t -4) is mulitple of 2 and hence even + 16 is even) and even + even = even...
1) t>1 , hence t^2 can be any number (mind you it can be a fraction as well) so answer is may or may not be ..NOT Sufficient
2) t is an even prime number ..i.e 2 so 2t is 4 and since the first expression and 16 both are divisible by 4 hence the sum of two numbers are also divisible by 4 ....Sufficient ...
hence the answer is B
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Re: Challenge - divisibility and exponents (m05q33)
[#permalink]
25 Oct 2012, 06:16
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