Suresh's Corner: Miscellaneous Questions From Quants Part-II

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Dear friends, this is the continuation of my very popular old thread. You can visit that thread here. http://www.pagalguy.com/forum/quantitative-questions-and-answers/10276-sureshs-corner-miscellaneous-questions-quants.html Unfortunat...
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Ok, here is my solution..

Given that LCM(x,y)=50! and GCD(x,y) = 5!

Let x = (5!)(a) and y = (5!)(b). (since GCD(x,y) = 51)
It is clear that a and b are co-primes.

LCM(x,y) = (5!)(a)(b) = 50!
Hence (a)(b) = 50!/5!

Now the number of possible values of a and b is the answer for the question asked.
Remember a and b are co-primes.
The number of ways in which a number N (which is 50!/5! in our case) can be written as product two co-primes is 2^(n-1) where n is the number of distinct prime factors of N.

Clearly 50!/5! will contain 15 distinct prime factors. (the number of primes less than 50)

Hence there are 2^14 combinations possible

I guess this is clear

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clsuresh1 Says
hmmmm, think again...I think you are trying to be too quick to conclude your answer...

In order to satisfy the given conditions either x or y(after excluding GCD ) has to be a prime number or product of prime numbers.

Now I am trying to fix one and then create other. First of all i take 0 prime number out of 12 possible values which can be done in 12C0 ways.
Then I take 1 prime number out of 12 which can be done 12C1 ways. The second number will be such that it satisfies the conditions.

Hence total no of ways = 12C0+12C1+12C2+12C3+.......+12C12

But in this only all the possible permutations also would have been considered. Hence answer should be 2^12.

Is the approach correct?
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shivam_01 Says
Can we also have the link Implex plzzz..

just read post number 63, few posts above this 😃
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implex Says
Suresh, your blog's content is awesome, i visit it almost daily!

Can we also have the link Implex plzzz..
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clsuresh1 Says
hmmmm, think again...I think you are trying to be too quick to conclude your answer...

Suresh, your blog's content is awesome, i visit it almost daily!
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Between 1 and 50 there are 15 prime numbers. And 3 between 1-5. Hence 12 are remaining. Total number of ways in which prime numbers can be selected = 2^12.

Hence total number of ordered pair = 2^13


hmmmm, think again...I think you are trying to be too quick to conclude your answer...
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Folks, I am back after a long gap.....
Anyway here is our next question.

How many ordered pairs of positive integers (x,y) exist such that LCM(x,y)=50! and GCD(x,y) = 5!

Between 1 and 50 there are 15 prime numbers. And 3 between 1-5. Hence 12 are remaining. Total number of ways in which prime numbers can be selected = 2^12.

Hence total number of ordered pair = 2^13
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Folks, I am back after a long gap.....
Anyway here is our next question.

How many ordered pairs of positive integers (x,y) exist such that LCM(x,y)=50! and GCD(x,y) = 5!

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Folks, well tried once again......The official answer is 70.

Here's the explanation.....

Since both 1000 and 2000 are of the form 2^px5^q, the numbers a,b,c must also be of this form

Let a=2^px5^m b=2^qx5^n c=2^rx5^s
Since LCM(a,b)=1000 LCM(b,c)=2000 and LCM(c,a)=2000
It is evident that max=3, max=4, max=4 ------------(1)
and max=3,max=3, max=3------------(2)

From(1) we must have r=4 and either p or q must be 3, while the other one can take the values 0,1,2 or 3.
Hence the number of tripplets (p,q,r) is 7
(0,3,4) (1,3,4) (2,3,4) (3,3,4)
(3,0,4) (3,1,4) (3,2,4)

From (2) two of m,n,s must be 3, while the third one can be 0,1,2 or3
Hence the number of tripplets (m,n,s) will be 10, namely
(3,3,4) (3,3,1) (3,3,2) (3,3,3)
(3,0,3) (3,1,3) (3,2,3)
(0,3,3) (1,3,3) (2,3,3)

Since the choice of (p,q,r) is independent of the choice of the (m,n,s), they can be chosen in 7x10 =70 ways.
Hence 70 ordered tripplets are possible.

I hope this is clear......

REgards,

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L(a,b) = 2^3* 5^3--------1
L(b,c) = 2^4 * 5^3-------2
L(c,a) = 2^4 * 5^3-------3


a/b has maximum power of 2 as 3. So for 2 and 3 to satisfy maximum power of 2 in c has to be 2^4(fixed).

Now for 1,2 and 3 to satisfy 2 of the three should have power of 5 as 5^3, whereas the third can take 4 values in terms of powers of 5(5^0 - 5^3). Hence total no of such arrangement = 3!/2! *4 = 12

Also either a or b should have 2^3 whereas other can take 4 vaues in terms of powers of 2(i.e- 2^0 to 2^3). Total no of such arrangements = 2*4=8

Combining both the conditions total no of such ordered pair should be 12*8 = 96.

Not Sure though :neutral:



Ok, my second attempt :-).

As selebratinglife has already pointed out that there will be some combinations repeating. Hence going on same line.

Either a or b should have 2^3. Hence fixing 1 as 2^3 other can take values from 2^0-2^3.
Hence total combination= 2*4 - 1(the one combination is 2^3,2^3 which we took twice)= 7

2 of the three should have power of 5 as 5^3, whereas the third can take 4 values in terms of powers of 5(5^0 - 5^3). Hence total no of such arrangement = 3!/2! *4 -2(this is for 5^3,5^3,5^3 which we took thrice) = 12 -2 =10

Hence answer should be 10*7 = 70

What do you say puys?
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