How many Integers from 0 to 100 are not divisible by 2,3,5 & 7

• there are 25 primes under 100,
  2,3,5,7, will be removed and 1 will be added hence ans =25-4+1=22
  or u can also do it by 100-AUBUCUB=100-(50+33+20+14)+(16+10+7+6+4+2)-(3+2+1)-0=22
• 7 years agoHelpfull: Yes(20) No(4)
• day after 2mrw is my elitmus test ..how should i prepare to score welll
• 7 years agoHelpfull: Yes(17) No(3)
• total no not divisible by 2,3,5,7 =total no - total no divisible by 2,3,5,7
  total no divisible by 2= last no - first no /gap +1
  ; no divisible by 2 =100-2/2 +1 =50
  ;no divisible by 3= 99-3/6+1= 17 (excluding no already divisible by 2)
  ;no divisible by 5 =5,25,55,65,85,95 =7 (excluding no already divisible by 2 &3)
  ; no divisible by 7= 7, 49,77,91 =4
  total no divisible by 2,3,5&7=78
  So total no not divisible =101-78=23
  So 23 is the total no of nos between 0 &100 not divisible by 2,3,5&7.
• 7 years agoHelpfull: Yes(6) No(1)
• Numbers divisible by 2 = 50
  
  Numbers divisible by 3 = 33
  
  Numbers divisible by 5 = 20
  
  Prime numbers from 0 to 100=25
  let's divide the problem into smaller domain= 0 to 10
  every number is divisible from 1 to 10= 2,3,,4,5,6,7,8,9,10
  answer is 21(25-4 (2,3,5,7))
• 7 years agoHelpfull: Yes(3) No(7)
• Div by 2=50
  Div by 3=33
  Div by 2&3=16
  So div by only 3=33-16=17
  Div by 5=20
  Div by 5&2&3=13
  So div by only 5=7
  Div only 7, not by 2,3,5=3
  Not div by 2,3,5,7=100-50-17-7-3=23
  I think this is correct
• 7 years agoHelpfull: Yes(3) No(1)
• 0-100 =101 num
  divisible by 2 , ;0,2,4...100 => 0+(d-1)2=100 =>d=51
  divisible by 3.; 3,9.15...99 => 3+(d-1)6=99 => d=17
  divisible by 5 .; 5 ,25 ,35 ,55,65,85 ,95 => total 7 num
  divisible by 7 .; 7 ,49 ,77 ,91 => total 4 num
  so toal divisible by 2, 3 5 , 7 from 0 to 100 => 51+17+7+4=79
  now , for we hv to find not divisible by them => 101-79 =22
  so ,22ans
• 7 years agoHelpfull: Yes(3) No(1)
• Every one submit different - different answer
  For check correct answer compile this code:
  #include
  int main(){
  int a;
  int count=0;
  printf("Numbers are:n");
  for(a=1;a
• 7 years agoHelpfull: Yes(1) No(1)
• we have 25 prime nos and every no is divisible by 1-9 except primes so we have 2 ,3 ,5 ,7 here but 1 is not divisible by those nos which is given in que so 25-- 4 (bcoz those 4 no are prime) = 21
  1 is also not divsble so 21+1 =22
  Ans is 22.
• 7 years agoHelpfull: Yes(1) No(0)
• lets start with numbers divisible by 2 3 5 7 and then subtracting it.
  divisible by 2 = 50 no.s (2x50=100)
  divisible by 3 = 33 no.s
  divisible by 5 = 20
  divisible by 7 = 14
  
  Now starting from above total numbers is 50+33+20+14=117
  now 3x33 has 16 even numbers added in divislbe by 2 = 117-16
  in divisible by 5, 10(10,20,30....) numbers are even so 117-16-10 and has 3 numbers divisible by 3 (15,45,75)
  in divisible by 7, 7 even and 2 numbers added in multiples of 3(21,63)
  so total = 117-16-10-3-7-2
  = 79 numbers divisible
  so 21 numbers not divisible
• 7 years agoHelpfull: Yes(0) No(7)
• 21
  and explanation is same as MOHAMMAD ANSAB
• 7 years agoHelpfull: Yes(0) No(4)