The integers between 1 and 100, inclusive, are put in list A if they a

Question

The integers between 1 and 100, inclusive, are put in list A if they are divisible by 2 and in list B if they are divisible by 3. How many integers in list A are not in list B?

(A) 11
(B) 16
(C) 25
(D) 33
(E) 34

bebs · Accepted Answer

# of items in list A =  \frac{100-2}{2}  +1 = 50. For list B, it is  \frac{99-3}{3}  +1 = 33. LCM of 2 and 3 is 6, factors of 6 ( \frac{96-6}{6}  +1 = 16) will be in list A and B ==> 50 - 16 = 34 are in A but not in B. The answer is   D

Archit3110 · Answer

A= 50 
B = 33
common set = LCM = 6 ; 16 
so A has 34 less not in B
IMO E

CaptainLevi · Answer

List A would have all the multiples of 2 i.e 2,4,6,8,10 and so on. Easy way to calculate multiples of a number within a range of numbers is \frac{(last multiple)-(first multiple)}{number} +1. \frac{100-2}{2} +1 = 50 Similarly with List B, all multiples of 3 and that would be 33 in size ( not required but in a hurry we tend go with the question stem and even get this number but we won't use it

) All the numbers - common numbers = Uncommon numbers common numbers= common multiples of 2 and 3 = 2X3 = 6 so multiples of 6. There are 16 multiples of 6 50 - 16 = 34 is the answer

yazlal25 · Answer

No. of terms in list A will be 50.

We have been asked how many of these 50 will not be in B.

Lets calculate how many of these will be in B. For a number to be divisible by both 2 and 3, it has to be divisible by 6.

Therefore common numbers in list A & B = factors of 6 between 1 & 100 = (96-6)/6 + 1 = 16

Therefore, numbers in list A not in list B = Numbers in list A - Numbers common in list A & B = 50 - 16 = 34 (E)

The integers between 1 and 100, inclusive, are put in list A if they a

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The integers between 1 and 100, inclusive, are put in list A if they a

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