22. How many even integers n, where 100  n  200, are divisible neither by seven nor by nine?

• first no. divisible by 7 = 112 (let the common diff be 14 (as we need even , keeping 7 will give odd) last even divisible by 7 is 196
  
  196 = 112+ (n-1)14
  
  n= 7
  
  
  similarly for the second
  
  
  198=108+18(n-1)
  
  n= 6
  
  
  so the number of even divisble = 6+7
  
  and non divisible = 50-13= 37
  
• 10 years agoHelpfull: Yes(24) No(8)
• to find n where 100 7+6-1 = 12
  therefore, the total no of even integers which are not divisible by 7 and 9 are -> 51-12= 39 (ans)
  
  
  
  
• 10 years agoHelpfull: Yes(17) No(7)
• the previous solution was wrong as I had considered all the nos.
  even integers between 200 and 100=51
  200=100+(n-1)2
  n=51
  no of even nos divisible by 7
  196=112+(n-1)14
  n=7
  no of even nos divisible by 9
  198=108+(n-1)18
  n=6
  no of even nos divisible by 63=1
  51-(6+7+1)=37
• 10 years agoHelpfull: Yes(8) No(2)
• First number 105 which is divisble by 7 between 100 to 200
  and 196 is the last which is divisible by 7,
  so number of numbers divisible by 7=(196-105)/7+1=14
  similarly for 9 it is (198-108)/9+1=11
  ans is=101-14-11=76
• 10 years agoHelpfull: Yes(7) No(7)
• divisible by 7=200/7-100/7=15(even no=7)
  divisible by 9=200/9-100/9=10(even no=5)
  total=12 ans
• 10 years agoHelpfull: Yes(2) No(6)
• @manofmayhem... a small rectification...51 even no.s from 100 to 200. 7+6-1 even no.s are divisible by 7 and 9.(-1 bcoz 126 is divisible by both 7 & 9).therefore ans=51-(7+6-1)=39
• 10 years agoHelpfull: Yes(2) No(4)
• total nos between 100 and 200=101
  nos divisible by 7=(196-105)/7+1=14
  nos divisible by 9=(198-108)/9+1=11
  nos divisible by both=2
  so, nos not divisible by either=101-(14+11-2)=78
• 10 years agoHelpfull: Yes(1) No(3)
• no. which is divisible by 7 in between 100 and 200 is-
  105,112,119,126,133,140,147,154,161,168,175,182,189,196.
  no divisible by 9 is-108,117,126,135,144,153,162,171,180,189,198
  126 and 189 is common then
  total term =12+9+2=23
  and 99-23=76
• 10 years agoHelpfull: Yes(0) No(2)
• even integers between 200 and 100=51
  200=100+(n-1)2
  n=51
  no of even nos divisible by 7
  196=112+(n-1)14
  n=7
  no of even nos divisible by 9
  198=108+(n-1)18
  n=6
  but two numbers are there 126 and 189 which are divisible by both 7 and 9.. so these two numbers are taken twice.
  thet is why 51-(7+6-2)=40
• 8 years agoHelpfull: Yes(0) No(1)
• 100
• 8 years agoHelpfull: Yes(0) No(1)
• even number between 100 and 200 = 49.
  even number divisible by 7 b/w 100 and 200 are (112,126,....,196).
  so, 196 = 112 + (n-1)*14
  => n = 7
  similarly, for 9.
  even number divisible by 9 b/w 100 and 200 are (108,126,....,198).
  so, 198 = 108 + (n-1)*18
  => n = 6
  
  total no div. by 7 or 9 are = 7+6 = 13
  
  but, here 126 is common in both AP's , so it is counted twice. so,
  total number div by 7 or 9 are 13-1 = 12.
  
  so, total number neither div. by 7 nor 9 are:
  
  49 - 12 = 37.
  
  (note: here we won't count 100 and 200 , because we need number between 100 and 200. so, we exclude them.)
• 7 years agoHelpfull: Yes(0) No(0)
• We can find this by removing numbers divisible by either 7 or 9 or both from all possible even numbers, where 100
• 6 years agoHelpfull: Yes(0) No(0)