Latest TCS Aptitude Question SOLUTION: in how many ways can the number 2310 be expressed as a product of 3 factors.

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in how many ways can the number 2310 be expressed as a product of 3 factors.

Asked In TCS harti (9 years ago)

Solved Narasapu Hari prasad Read Solution (23)

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Read Solution (23) :

Aniruddha  (9 years ago)

2310=2*3*5*7*11
two possibilities (as per question):
grouping 3 factors at once => 5c3*1*1 (eg 30*7*11)= 10
grouping 2 factors, 2 factors and 1 factor => 5c2*3c2*1 (eg 6*35*11) = 30

so 30+10 = 40 ways !

Is this solution Helpfull? Yes (20) | No (8)  

Narasapu Hari prasad  (9 years ago)

Ans is 243

2310=11*7*3*5*2

a=11^p1*7^q1*3^r1*5^s1*2^t1.
B=11^p2*7^q2*3^r2*5^s2*2^t2.
C=11^p3*7^q3*3^r3*5^s3*2^t3.
Where a*b*c=2310 and
p1+p2+p3=1
q1+q2+q3=1
r1+r2+r3=1
s1+s2+s3=1
t1+t2+t3=1

So 3*3*3*3*3=243

Is this solution Helpfull? Yes (10) | No (21)  

aleem  (9 years ago)

We have 2310 = 2*3*5*7*11

Say 2310 = _ * _ * _

Now, each number can go at 3 places. So, we will get 3^5 i.e. 243 ways.

But, 2310*1*1, 1*2310*1 and 1*1*2310 are same. So, we need to remove 2 cases out of 3 here. Out of remaining 240 cases, we need to consider 240/(3!) i.e. 40 cases.

So, total should be 41.

Is this solution Helpfull? Yes (9) | No (1)  

chakri  (9 years ago)

{3 Factors}*{1 Factor}*{ 1 Factor}
or {2 Factors}*{2 Factors}*{ 1 Factor}

5c3 + 5c2*3c2 = 40

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sweety  (9 years ago)

41 is the correct ans
http://www.pagalguy.com/cat/official-quant-thread-for-cat-2011-part-4-25070678/6543045

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poulami   (9 years ago)

ans 5C3 = 10

Is this solution Helpfull? Yes (2) | No (6)  

Swapnil Dubey  (9 years ago)

We have 2310 = 2*3*5*7*11

Say 2310 = _ * _ * _

Now, each number can go at 3 places. So, we will get 3^5 i.e. 243 ways.

But, 2310*1*1, 1*2310*1 and 1*1*2310 are same. So, we need to remove 2 cases out of 3 here. Out of remaining 240 cases, we need to consider 240/(3!) i.e. 40 cases.

So, total should be 41.

Is this solution Helpfull? Yes (2) | No   

Sreelatha  (9 years ago)

The correct must be 41..

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sheela kumari singh  (9 years ago)

2310 = 2*3*5*7*11
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3^(n-1) + 1)/2 ways=41 ways

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kus  (9 years ago)

THIS SITE SUCKS IN SOLUTIONS!!!!

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vivek kumar  (9 years ago)

num of ways= 3*3!=18

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surya teja penigalapati  (9 years ago)

ans:18
sol:3*3*2*1=18

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Rahul  (9 years ago)

4

ans:4c3

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Narasapu Hari prasad  (9 years ago)

Ans 15. If valus of factors are not same.
5c1+5c2=5+10=15

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rahul gera  (9 years ago)

5C3 ways
=10

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Aditya keri  (9 years ago)

2310 = 2*3*5*7*11....
A product of 1st 5 prime numbers... 5 Factors!


U need to take the product of 3 factors out of these 5 factors....
Which can be taken in two ways.....
{1 Factor}*{1 Factor}*{the rest 3 Factors}
or {1 Factor}*{2 Factors}*{the rest 2 Factors}

1st Case: There are 5*4 ways = 20 Ways
2nd Case: There are 5*6(4C2) ways = 30 Ways

A total of 50 ways in which 2310 can be expressed as a product of three factors....

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mounika  (9 years ago)

how can you write 5*4,5*4(6c2)ways Mr.aditya

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Aashish Sawant  (9 years ago)

2310=2*3*5*7*11;
mulitple of three=(2*3)(5*7)(11)
(2*3)(5*11)(7);
like this for other number so..30 ways...

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shashank shekhar   (9 years ago)

2310=2*3*5*7*11
here is two conditions if we club these factors in to multiple of three factors.that are 2.2.1 and 1.1.3
number of ways =(5c2.4c2.1)+(5c1.4c1.3)=120

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sahil shukla  (9 years ago)

correct ans =14

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soumya  (9 years ago)

can anyone tell me what is the correct solution for this ????

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aman sharma  (9 years ago)

2310=_*_*_ each have 5 choices so 5^3
that cant be true
bcz it makes 2310=2*3*5 or any case n which is not true

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ANKIT PATEL  (8 years ago)

2310=2*3*5*5*11
then total possible ways for expressed 2301 as a multiple of three numbers is 4p3=24

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