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# Three questions from the gmat prep

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Three questions from the gmat prep [#permalink]

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28 May 2009, 21:42
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EDIT: 1 more question added, see my last post at the end of this page

Hi guys,

I need help with the following three questions as i really have no clue on how to derive the answers. I greatly appreciate your help.

1. What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11

2. attached
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1.gif [ 11.39 KiB | Viewed 1678 times ]

3. attached
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Untitled-2.gif [ 1.18 KiB | Viewed 1680 times ]

Last edited by steliossb on 29 May 2009, 09:01, edited 2 times in total.
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Re: Three questions from the gmat prep [#permalink]

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28 May 2009, 23:00
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the first question has been recently discussed here. You can do a search for that one.

I'll take a crack at the other two

problem 2.

let q and m=quotients

given:

n=3q+2, t=5m+3

what is the remainder when nt/15?

nt/15= (3q+2)(5m+3)/15=

(15qm+9q+10m+6)/15

1) (n-2)/5 is an integer

n=3q+2 so 3q/5 is an integer

since 3 and 5 are prime q must be a multiple of 5

plugging into our equation 15qm+9q+10m+6)/15, we see that the first and second terms are multiples of 15, but we dont no about the third term

so insufficient

2) t/3 is an integer so t is a multiple of 3

t=5m+3 so m must be a multiple of 3 as well

plugging into our equation 15qm+9q+10m+6)/15 we find that the first and third terms are multiples of 15 but we dont know about the second term

so insufficient

togther we know all terms are multiples of 15 and the remainder is 6
sufficient

C

problem 3.

if x is -tive then whats the sqrt of -xlxl ?

if x is -tive then -x is x and lxl is also x so the sqrt of x^2 is x

so D
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 01:51
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(1) Prime factor is 2.

4^17 - 2^28 = 2^34 - 2^28 = 2^6 = 64
So 2 is the only prime factor for 64

(2) Answer is C, we can derive the conclusion only with both the statements

(3) Since X < 0, it has to be a -ve number.
If X = -2
So, substitute -2 in the equation
Sqrt (-(-2)|-2|) = Sqrt (2*2) = 2

We took X = -2 and got an answer as 2, so the value will be -X
So I would go for -X, Option A

Thanks,
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 05:37
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vesam wrote:
(1) Prime factor is 2.

4^17 - 2^28 = 2^34 - 2^28 = 2^6 = 64
So 2 is the only prime factor for 64

You can't add or subtract exponential power in addition and subtraction of bases.

Powers would be added or subtracted in case of multiplication and division only.

4^17 - 2^28 = 4^17 - 4^14 = 4^14 ( 4^3 - 1) = 4^14*63 = 4^14*7*9.

Greatest prime factor = 7

Last edited by humans on 29 May 2009, 09:58, edited 1 time in total.
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 06:59
I misread the question.
Thanks for the correct answer.
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 07:29
thanks a lot for the help guys, it was extremely useful.

last question i have for these forums:

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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 10:18
steliossb wrote:
thanks a lot for the help guys, it was extremely useful.

last question i have for these forums:

This question is already answered here gmatprep-practice-question-probability-coin-77577.html
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 12:11
humans wrote:
steliossb wrote:
thanks a lot for the help guys, it was extremely useful.

last question i have for these forums:

This question is already answered here gmatprep-practice-question-probability-coin-77577.html

There is an easier approach/formula for such problems rather than working out all the possible combinations.

If the probability of success in each trial is p, then the probability of r successes in n trials is $$^nC_r$$ $$p^r$$ $$q^{n-r}$$

Since, the question asks us to find out probability for at least 4 heads, it gives rise to two scenarios. r = 4 heads exactly or r = 5 heads exactly. And ADD the scenarios finally.

We can infer from the question that probability of heads = p = 0.6, r = 4 or 5, n=5 . Obviously probability for tails = q = 0.4.

Substituting
$$^5C_4$$ $$(0.6)^4$$ $$(0.4)^{5-4}$$ + $$^5C_5$$ $$(0.6)^5$$ $$(0.4)^{5-5}$$

= $$5 (0.6)^4 (0.4) + (0.6)^5$$

Solving the above we get the answer E
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Re: Three questions from the gmat prep [#permalink]

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29 May 2009, 14:43
vesam wrote:
(1) Prime factor is 2.

4^17 - 2^28 = 2^34 - 2^28 = 2^6 = 64
So 2 is the only prime factor for 64

(2) Answer is C, we can derive the conclusion only with both the statements

(3) Since X < 0, it has to be a -ve number.
If X = -2
So, substitute -2 in the equation
Sqrt (-(-2)|-2|) = Sqrt (2*2) = 2

We took X = -2 and got an answer as 2, so the value will be -X
So I would go for -X, Option A

Thanks,

oops, you're right, should be -x, my bad
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Re: Three questions from the gmat prep [#permalink]

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19 Dec 2010, 15:45
On question 3, I got answer as D as well. sqrt(x*-x) or sqrt(-x^2)=x right? Please advise. thanks.
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Re: Three questions from the gmat prep [#permalink]

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19 Dec 2010, 16:06
gettinit wrote:
On question 3, I got answer as D as well. sqrt(x*-x) or sqrt(-x^2)=x right? Please advise. thanks.

The first question: factors-104757.html
The second question: gmat-prep-2-remainder-86155.html
The third question: square-root-and-modulus-100303.html (OA: A)
The fourth question: probability-question-gmatprep-85802.html

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Re: Three questions from the gmat prep   [#permalink] 19 Dec 2010, 16:06
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