How many positive integers between 200 and 300 (both inclusive) are no

Question

How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3
B. 16
C. 24
D. 26
E. 75

4d · Accepted Answer

Divisible by 2 =  \frac{300-200}{2} +1 = 51 
Divisible by 3 =  \frac{300-201}{2} +1 = 34 
Divisible by 5 =  \frac{300-200}{5} +1 = 21 
Divisible by 2 and 3 = \frac{300-204}{6} +1 = 17 
Divisible by 3 and 5 =  \frac{300-210}{15} +1 = 7 
Divisible by 5 and 2 =  \frac{300-200}{10} +1 = 11 
Divisible by 2,3 and 5 =  \frac{300-210}{30} +1 = 4 
positive integers that are not divisible by 2, 3 or 5 =  101 - (51+34+21) + (17+7+11) - 4 = 26 
IMO D

KanishkM · Answer

IMO D

Total numbers will be 101 (300-200+1)

Not divisible will be, if we can find the numbers divisible by 2, 3 or 5, we can subtract it from Total number

Since the terms will be in an AP, we can use  a_n = a + (n-1)d

Divisible by 2 = 300-200/2 +1 =51

Divisible by 3 = 300-201/3 + 1 = 34

Divisible by 5 = 300-200/5 + 1 = 21

Divisible by 2 or 3 = 300-204/6 + 1 = 17

Divisible by 2 or 5 = 300-200/10 + 1 = 11

Divisible by 3 or 5 = 300-210/15 + 1 = 7

Divisible by 2 ,3 or 5 = 300-210/30 + 1 = 4

101 - (Divisible by all 3) + (common terms of 2) - ( Common terms of all three) 
101 -( 51 + 34+21) + (17+11+7) - 4
136- 110
26

Archit3110 · Answer

total integers = 300-200+1 ; 101
+ve integers divisible by 2 = 51
integers divisible by 3 = 34
integers divisible by 5 = 21

integers divisible by both 2& 3 = half of integers divisible by 3 ' ie even integers divisible by 3' ; 17
integers divisible by both 2& 5 = half of integers divisible by 5 ' ie even integers divisible by 5'; 11
integers divisible by 3 & 5 both = 225,255,285 ; 3

so total integers divisible by 2,3 & 5 = (51-17-11)+ ( 34-3) + 21 = 75
integers not divisible by 2,3,5 would be 101- 75 = 26 
IMO D

MathRevolution · Answer

Total numbers: 300 - 200 + 1 = 101

Total numbers divisible by 2: First   and Last

=> 300 = 200 + (n-1)2
=> 102 = 2n
=> n = 51

Total numbers divisible by 3: First   and Last

=> 300 = 201 + (n-1)3
=> 102 = 3n
=> n = 34

Total numbers divisible by 5: First   and Last

=> 300 = 200 + (n-1)5
=> 105 = 5n
=> n = 21

Total numbers divisible by 6 : First   and Last

=> 300 = 204 + (n-1)6
=> 102 = 6n
=> n = 17

Total numbers divisible by 6 : First   and Last

=> 300 = 210 + (n-1)15
=> 105 = 15n
=> n = 7

Total numbers divisible by 10 : First   and Last

=> 300 = 200 + (n-1)10
=> 110 = 10n
=> n = 11

Total numbers divisible by 30 : First   and Last

=> 300 = 210 + (n-1)30
=> 120 = 30n
=> n = 4

Total numbers not divisible by '2' , '3' or '5': 101 - 51 - 34 - 21 + 17 + 7 + 11 - 4 = 26

Answer D

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How many positive integers between 200 and 300 (both inclusive) are no

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