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Difficulty:
55%
(hard)
Question Stats:
59% (01:37) correct
41%
(01:41)
wrong
based on 337
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How many positive integers can be expressed as a product of two or more of the prime numbers 5, 7, 11, and 13 if no one product is to include the same prime factor more than once?
A. Eight
B. Nine
C. Ten
D. Eleven
E. Twelve
_________________
A. Eight
B. Nine
C. Ten
D. Eleven
E. Twelve
_________________
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Bunuel
Two ways
1) combination..
Therefore, we have to add ways to pick 2 or 3 or all out of 4, so 4C2+4C3+4C4 = 6+4+1 = 11
2) there are two ways for each number - select or not select..
Therefore total ways is 2*2*2*2=16
Now this includes the 4 ways when each is taken alone, so subtract 4 from 16 = 16-4=12
It also includes the way when none is selected, so subtract 1..12-1=11
D
_________________
Bunuel
Notice the word two or more factors.
Selecting 2 primes out of given 4= 4C2 ways= 6
Selecting 3 primes out of given 4= 4C3 ways =4
Selecting 4 primes out of given 4= 4C4 ways= 1
So total ways = 6+4+1 = Eleven
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Bunuel
total given prime no 5,7,11,13
no of different products which can be made using these 4 prime no and not including same prime factor >1 is
4c2+4c3+4c4
6+4+1 = 11
IMO D
you can also calculate
5*7
5*11
5*13
7*11
7*13
11*13
5*7*11
7*11*13
11*13*5
13*5*7
5*7*11*13
5*7
5*11
5*13
7*11
7*13
11*13
5*7*11
7*11*13
11*13*5
13*5*7
5*7*11*13
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