Problem Solving

Q:In how many ways can 2310 be expressed as a product of 3 factors ?

A)21

B)23

C)41

D)46

E)56

Aman verma wrote:Q:In how many ways can 2310 be expressed as a product of 3 factors ?

A)21

B)23

C)41

D)46

E)56

Let there be 3 "factor boxes" A, B and C.

2310 = 2*3*5*7*11.
To yield a product of 2310, each of the red values above must be placed in factor box A, B or C
Note:
If a factor box is left empty, the value of that factor box = 1.

Since there are 3 options for each of the 5 red values, the number of possible distributions = 3*3*3*3*3 = 243.

BUT:
Among these 243 distributions, many are composed of the same 3 factors:
1*1*2310 represents the same set of factors as 1*2310*1.
To avoid over-counting, we must account for the DUPLICATE SETS of factors.

Option 1: 2 distinct factors
Here, the factors must be 1, 1 and 2310, implying 3 cases:
1*1*2310
1*2310*1
2310*1*1.
Only 1 of these 3 cases must be included in our total.
Total ways = 1.

Option 2: 3 distinct factors
Since 3 of the 243 cases are composed of 2 distinct factors, the number of cases with 3 distinct factors = 243-3 = 240.
Among these 240 cases, any arrangement of the 3 distinct factors represents the same set of factors:
1*30*77 is the same set of factors as 77*1*30.
Thus, to avoid over-counting, we must be divide by the number of ways the 3 distinct factors can be ARRANGED (3!):
Total ways = 240/3! = 40.

Result:
Number of ways to yield 3 factors = 1+40 = 41.

The correct answer is C.

GMATGuruNY wrote:

Aman verma wrote:Q:In how many ways can 2310 be expressed as a product of 3 factors ?

A)21

B)23

C)41

D)46

E)56

Let there be 3 "factor boxes" A, B and C.

2310 = 2*3*5*7*11.
To yield a product of 2310, each of the red values above must be placed in factor box A, B or C
Note:
If a factor box is left empty, the value of that factor box = 1.

Since there are 3 options for each of the 5 red values, the number of possible distributions = 3*3*3*3*3 = 243.

BUT:
Among these 243 distributions, many are composed of the same 3 factors:
1*1*2310 represents the same set of factors as 1*2310*1.
To avoid over-counting, we must account for the DUPLICATE SETS of factors.

Option 1: 2 distinct factors
Here, the factors must be 1, 1 and 2310, implying 3 cases:
1*1*2310
1*2310*1
2310*1*1.
Only 1 of these 3 cases must be included in our total.
Total ways = 1.

Option 2: 3 distinct factors
Since 3 of the 243 cases are composed of 2 distinct factors, the number of cases with 3 distinct factors = 243-3 = 240.
Among these 240 cases, any arrangement of the 3 distinct factors represents the same set of factors:
1*30*77 is the same set of factors as 77*1*30.
Thus, to avoid over-counting, we must be divide by the number of ways the 3 distinct factors can be ARRANGED (3!):
Total ways = 240/3! = 40.

Result:
Number of ways to yield 3 factors = 1+40 = 41.

The correct answer is C.

Hi Mitch,

Can you please explain it in a simple way of counting.
I think we can use combinations to solve it.

we know prime factors are 2,3,5,7

nikhilgmat31 wrote:Hi Mitch,

Can you please explain it in a simple way of counting.

Ignore this problem.
It is too complex and time-consuming for the GMAT.

GMATGuruNY wrote:

nikhilgmat31 wrote:Hi Mitch,

Can you please explain it in a simple way of counting.

Ignore this problem.
It is too complex and time-consuming for the GMAT.

Is it a real GMAT question ?