TCS
Company
undefined
in how many ways can the number 2310 be expressed as a product of 3 factors.
Read Solution (Total 23)
-
- 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 ! - 10 years agoHelpfull: Yes(20) No(8)
- 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 - 10 years agoHelpfull: Yes(10) No(21)
- 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. - 10 years agoHelpfull: Yes(9) No(1)
- {3 Factors}*{1 Factor}*{ 1 Factor}
or {2 Factors}*{2 Factors}*{ 1 Factor}
5c3 + 5c2*3c2 = 40 - 10 years agoHelpfull: Yes(8) No(0)
- 41 is the correct ans
http://www.pagalguy.com/cat/official-quant-thread-for-cat-2011-part-4-25070678/6543045 - 10 years agoHelpfull: Yes(3) No(0)
- ans 5C3 = 10
- 10 years agoHelpfull: Yes(2) No(6)
- 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. - 10 years agoHelpfull: Yes(2) No(0)
- The correct must be 41..
- 10 years agoHelpfull: Yes(1) No(0)
- 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
- 10 years agoHelpfull: Yes(1) No(0)
- THIS SITE SUCKS IN SOLUTIONS!!!!
- 9 years agoHelpfull: Yes(1) No(1)
- num of ways= 3*3!=18
- 10 years agoHelpfull: Yes(0) No(1)
- ans:18
sol:3*3*2*1=18 - 10 years agoHelpfull: Yes(0) No(5)
- 4
ans:4c3 - 10 years agoHelpfull: Yes(0) No(1)
- Ans 15. If valus of factors are not same.
5c1+5c2=5+10=15 - 10 years agoHelpfull: Yes(0) No(5)
- 5C3 ways
=10 - 10 years agoHelpfull: Yes(0) No(2)
- 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.... - 10 years agoHelpfull: Yes(0) No(5)
- how can you write 5*4,5*4(6c2)ways Mr.aditya
- 10 years agoHelpfull: Yes(0) No(2)
- 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... - 10 years agoHelpfull: Yes(0) No(1)
- 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 - 10 years agoHelpfull: Yes(0) No(4)
- correct ans =14
- 10 years agoHelpfull: Yes(0) No(1)
- can anyone tell me what is the correct solution for this ????
- 10 years agoHelpfull: Yes(0) No(0)
- 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 - 10 years agoHelpfull: Yes(0) No(0)
- 2310=2*3*5*5*11
then total possible ways for expressed 2301 as a multiple of three numbers is 4p3=24 - 9 years agoHelpfull: Yes(0) No(0)
TCS Other Question