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Some problems: Number properties

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Some problems: Number properties

by GmatTakerNo.1 » Sat Apr 24, 2010 8:35 am
Hey,

I´ve got some problem with these exercises. Please, can you solve it and explain the way how you tackle these problem at best?

1. What is the highest power of 14 that divides 56! evenly?

Answer: 9


2. What is the highest integer power of 12 that divides 27! evenly?

Answer: 11


3. Which is bigger 89! - 88! or 87! × 88^2?

Answer: They are equal.

4. If 17! + 2 ≤ K ≤ 17! + 17, is K ever a prime number?

Answer: No never


5. If n is the product of the integers from1 to 20 inclusive, which of the following is the greatest integer k for which 2k is a factor of n?
a) 408
b) 437
c) 486
d) 532
e) 1242

Anser: D
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by iamseer » Sat Apr 24, 2010 10:07 am
GmatTakerNo.1 wrote:Hey,

I´ve got some problem with these exercises. Please, can you solve it and explain the way how you tackle these problem at best?

1. What is the highest power of 14 that divides 56! evenly?

Answer: 9
56!=1*2*3....55*56
14=2*7
So basically you have to find 14s in the factors of 56!
2*7, 14, 21 has 7 and 2 can be obtained from any even number, similarly for 28,35,42,56 and 49 is special b'cos it has two 7s (thats a GMAT trap :) ) So in all nine 14s can be formed.
GmatTakerNo.1 wrote: 2. What is the highest integer power of 12 that divides 27! evenly?

Answer: 11
27!=1*2*3...27
You have to find 12s in the 27!
Don't know if there is a smarter way. I do it the old fashioned way of writing all the prime factors of numbers upto 27 and then finding all the 12s. Turns out you can make eleven 12s.
GmatTakerNo.1 wrote:
3. Which is bigger 89! - 88! or 87! × 88^2?

Answer: They are equal.
This one is simple.
89!-88!=88!(89-1)=88!*88=87!*88^2
GmatTakerNo.1 wrote: 4. If 17! + 2 ≤ K ≤ 17! + 17, is K ever a prime number?

Answer: No never
This one looks interesting :)
17! is a even number so u can be sure that 17! + any even number will be even and hence not prime
now for number like 17!+3 = 3(1*2*4*...*16*17+1)=3(even+1) a multiple of 3
similarly 17!+5 is 5(1*2*4*6...*16*17+1)=5(even+1) amultiple of 5
similarly for 7,9,11,13,15,17
therefore No never
GmatTakerNo.1 wrote:
5. If n is the product of the integers from1 to 20 inclusive, which of the following is the greatest integer k for which 2k is a factor of n?
a) 408
b) 437
c) 486
d) 532
e) 1242

Anser: D
they have asked for greatest so start with greatest value in answer option
1242=2*3*3*3*23 .... not our guy b'cos it has a prime factor greater than 20
lets try the next one
532=
=2*266
=2*2*133 (here i stopped for a moment b'cos 133 looked prime. When such things happen, we try bigger prime numbers, so I tried 7 and yes it works)
=2*2*7*19
this is our guy
"Choose to chance the rapids and dance the tides"
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by GmatTakerNo.1 » Sat Apr 24, 2010 11:07 am
Hey, thank you for your explanations. I understand everything.

Just another question, is there maybe another approach to solve problem 2?
The old fashioned way took me around 2 and a half to three minutes. Maybe a little bit too long during the real test.
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by GmatTakerNo.1 » Sat Apr 24, 2010 11:40 am
I got some more:

6. For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is?
a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 50

The answer is E.

And how can I get a solution of these under two minutes?

7. What is the least common multiple of 8, 9, 10, 11, 12 and 24?

Answer: 3960

8. If a marching band has 72 members that always march in formations of at least three rows and at least 3 members in each row, how many different formations can they march in?

Answer: 8.

9. If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?

Answer: 14


10. The product of the first twelve positive integers is divisible by all of the following EXCEPT?

Answer: E
Thanks for your help in advance.
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by this_time_i_will » Sun Apr 25, 2010 1:14 am
GmatTakerNo.1 wrote:I got some more:

6. For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is?
a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 50

The answer is E.

.
h(100) = h(2*4*6*8*....*100) = h( (2^50)*(1*2*......*50)) = h(2^50 * 50!). so all the prime number from 2 to 50, would divide h(2^50 * 50!). And all these prime numbers between 2 and 50 would NOT divide h(2^50 * 50!) + 1.
They would always leave a reminder of 1. (And intresting point to note here is that this logic would not be applicable for h(2^50 * 50!) + 2 or h(2^50 * 50!) + 3 or h(2^50 * 50!) + 5 or h(2^50 * 50!) + 11 , since all these would be divisble by a prime number included in 50!). So the required prime number would be greater than 50.
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by this_time_i_will » Sun Apr 25, 2010 1:26 am
GmatTakerNo.1 wrote:I got some more:

And how can I get a solution of these under two minutes?

7. What is the least common multiple of 8, 9, 10, 11, 12 and 24?
u need to calculate LCM. If answer choices are given then a quick strategy may be thought of.
Answer: 3960

8. If a marching band has 72 members that always march in formations of at least three rows and at least 3 members in each row, how many different formations can they march in?
let the marching order is represnted by R*C, for Rows and Columns.
72 = 3^2*2^3.
Take one 3 for Rows and other for column from 3^2.
so [3]*[2^3*3].....the contents of first bracket indicates rows and that of second indicates column.
Now you can not shuffle 3 between these two bracket. You can only shuffle three 2's.Answer: 8.

9. If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?
Any number can be expressed as product of prime numbers p^a* q^b* r^c, where p,q,r etc. are prime numbers.
And, total number of factors, including 1 and number, is given by = (a+1)(b+1)(c+1).....
So, for this problem we have= p^a* q^b* r^c*s^d, where a=b=c=d=1.
total number of factors = (1+1)*(1+1)*(1+1)*(1+1) = 16.
So excluding 1 and number = 14
Answer: 14


10. The product of the first twelve positive integers is divisible by all of the following EXCEPT?

Answer: E
Thanks for your help in advance.
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by Testluv » Sun Apr 25, 2010 1:29 am
GMATtaker,

welcome to the site.

Please observe two rules to help us in keeping things organized:

1) please post only one question per thread;

2) always use the search function (upper right corner of the screen) prior to posting. Almost all of the questions you've posted have been discussed (in some cases, a lot more than once). If you use the search funtion, you can automatically find pages of solutions to most of the questions you've posted.
Kaplan Teacher in Toronto
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by GmatTakerNo.1 » Sun Apr 25, 2010 3:16 am
Thanks for the inffo about the search function.

One more question: What does LCM stand for as in the answer of "This time i will"?

Thanks.
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by eaakbari » Wed Apr 28, 2010 5:11 am
iamseer wrote:here i stopped for a moment b'cos 133 looked prime. When such things happen, we try bigger prime numbers, so I tried 7 and yes it works
Use the divisibility tests, instead of just trying bigger numbers

https://en.wikipedia.org/wiki/Divisibility_rule

HTH
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by eaakbari » Wed Apr 28, 2010 5:45 am
GmatTakerNo.1 wrote:

2. What is the highest integer power of 12 that divides 27! evenly?

Answer: 11
This question is not as complicated as it seems. In general in these type of question, factorize our divisor
12=2*2*3.
2*2=4
Now Since 4 will appear less often than a 3, lets find the number of 4's by first finding the number of 2's

In general the formula for finding the number of k's in a number m!
m/k + m/(k^2) + m/(k^3)....m/(k^n)
so on and so forth where k^n<m

So for 2 , we get
27/2 + 27/4 +27/8 + 27/16

13+6+3+1 = 23

Now 27! has 23 2's
Hence the number of 4's is 23/2 = 11

Hence Answer 11
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by iamseer » Wed Apr 28, 2010 6:07 am
eaakbari wrote:
In general the formula for finding the number of k's in a number m!
m/k + m/(k^2) + m/(k^3)....m/(k^n)
so on and so forth where k^n<m

So for 2 , we get
27/2 + 27/4 +27/8 + 27/16

13+6+3+1 = 23

Now 27! has 23 2's
Hence the number of 4's is 23/2 = 11

Hence Answer 11
Thanks for sharing this formula.
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