20. How many numbers are there from 100 to 1200 which are not divisible either by 2, 3, or 5?

• tn=a+(n-1)d
  For 2=>n=551
  For 3=>n=367
  For 5=>n=221
  For (2*3)=6=>n=184
  For (3*5)=15=>n=74
  For (5*2)=10=>n=111
  For LCM (2,3,5)=30=>n=37
  So NUMBERS are=(551+367+221)-(184+74+111)+37
  =1139-369+37
  =807
  
  Therefore no of numbers not divisible=1200-99-807=294
• 11 years agoHelpfull: Yes(54) No(4)
• Sn=a+(n-1)d
  For 2=>n=550
  For 3=>n=267
  For 5=>n=220
  But in 5 Half of the terms will end with 0 hence they are also divisible by 2
  So 1098-(550+367+110)=1098-1027=71
• 11 years agoHelpfull: Yes(1) No(8)
• tn=a+(n-1)d
  For 2=>n=551
  For 3=>n=367
  For 5=>n=221
  For (2*3)=6=>n=184
  For (3*5)=15=>n=74
  For (5*2)=10=>n=111
  For LCM (2,3,5)=30=>n=37
  So NUMBERS are=(551+367+221)-(184+74+111)+37
  =1139-369+37
  =807
• 11 years agoHelpfull: Yes(1) No(4)
• 1/2 of the no.s divisible by 2 so 550 no.s divisible and for5 as already cancelled the
  10 20 30 40........by 2 so in every 10no.s one will be divisible for 55 tens. 55 can divisible
  So 550-55=495
  Then after removing 2 and 5 multiples in every 30 no.s 4 will divisible
  For 495 no.s 16*4+2=66
  So ans is 495-66=429
• 11 years agoHelpfull: Yes(1) No(0)
• 294 answer
• 11 years agoHelpfull: Yes(1) No(0)
• what is 1098 here
• 11 years agoHelpfull: Yes(0) No(0)
• Priyanka..1098 is the total no. of terms from a=100 and tn=1200..
  
• 11 years agoHelpfull: Yes(0) No(3)
• ans--1139...
• 11 years agoHelpfull: Yes(0) No(2)
• n(AUBUC) = n(A) + n(B) + n(C) - n(A intersectn B) - n(B intersectn C) - n(A intersectn C) + n(A intersectn B intersectn C)
  
  n(2U3U5) = n(2) + n(3) + n(5) - n(6) - n(15) - n(10) + n(30)
  
  --> n(2U3U5) = 600 + 400 + 240 - 200 - 80 - 120 + 40 = 880
  
  Hence number of numbers not divisible by any of the numbers 2, 3, and 5
  
  = 1100 - 880 = 220.
• 11 years agoHelpfull: Yes(0) No(4)
• total number of terms b/w 100 and 1200 are 1001
  out of it 501 are even
  remaining terms=500
  numbers divisible by 3
  3*34=102
  3*400=1200
  so numbers divisible by 3 are 400-34+1=367
  remaining terms=500-367=133
  similarly number divisible by 5 221
  remainig number 133-221= -88
  now adding multiples of 2*3(6),2*5(10),3*5(15) as they are reapetely subtracted
  ie
  we get 281 as total number not divisible by 2, 3,or 5
  
  
• 11 years agoHelpfull: Yes(0) No(0)
• why - 99 from 1200??
• 8 years agoHelpfull: Yes(0) No(0)