Find the number of number between 100 to 400 which are divisible by either 2 ,3 ,5 and 7

• no. divisible by 2 are 151(including 100 &400)
  no. divisible by 5 but not by 2 are 30
  no. divisible by 3 but not by 2 and 5 are 41
  no. divisible by 7 but by 2 ,5 and 7 are 11
  therefore, total no. are 233
  
• 9 years agoHelpfull: Yes(15) No(5)
• The difference between 100 and 400 is 300.
  So,there will be 151 no. which will be divisible by 2(If you divide 300 by 2 you will get 150 and if you consider 400 then total 151.).
  So,there will be 101 no. which will be divisible by 3.
  So,there will be 61 no. which will be divisible by 5.
  So,there will be 42 no. which will be divisible by 7.
  and therefore total 355 no which will be divisible by either 2 or 3 or 5 or 7.
• 9 years agoHelpfull: Yes(10) No(14)
• no. divisible by 2 are 151
  no. divisible by 3 but not by 3 are 50
  no. divisible by 5 but not by 2 and 3 are 20
  no. divisible by 7 but by 2 ,3,5 are 11
  therefore, total no. are 232
• 8 years agoHelpfull: Yes(10) No(0)
• I have a simpler way. Let's find the number of numbers not divisible by 2,3,5,7.
  
  First, there are 150 numbers divisible by 2. Take them away. Of the remaining, every third one is divisible by 3. Take them out and there are 100 left. Every 5th is divisible by 5. 100/5 is 20 so when taken out there are 80 left. Every 7th is divisible by 7, so there remain 6/7 of these: 80*6/7=68.57. Round down to 68.
  
  So we have 300-68=232 that are divisible by 2,3,5,7.
• 8 years agoHelpfull: Yes(9) No(0)
• solving this ques by ap series
  ap series of no divisible by 2 102,104,106.....398 (as the ques is number in between 100 t0 400)
  
  by applying the nth term fornula
  t(n)=a+(n-1)d where a=102, d=2 ,t(n)=398
  398=102+(n-1)2
  398=102+2n-2
  298=2n
  n=149
  and this can be done for other numbers as well... and btw it has asked about no of numbers not sum of numbers.
• 8 years agoHelpfull: Yes(2) No(2)
• LET of 2 3 5 7= 210
  Ans is 210 numbers
• 9 years agoHelpfull: Yes(1) No(16)
• 1 as take lcm of 2,3,5,7 that is 210
• 9 years agoHelpfull: Yes(1) No(8)
• ans is 231... i manually calculate the no.
  
• 9 years agoHelpfull: Yes(1) No(2)
• the lcm of 2,3,5,7 is 210 so only one is answer
• 9 years agoHelpfull: Yes(1) No(6)
• no divisible by 2 are 102,104,106.....398 (as the ques is number in between 100 t0 400)
  
  by applying the nth term formula of AP
  t(n)=a+(n-1)d where a=102, d=2 ,t(n)=398
  398=102+(n-1)2
  398=102+2n-2
  298=2n
  n=149
  no divisible by 3 are 102,105,108,....................,399
  by applying the nth term formula of AP a=102 , d=3 ,t(n)=399
  399=102+(n-1)3
  we get n=100
  no. which are divisible by 2 and 3 both are = 102,108,114......................,396
  so these are=50 and these 50 no. are included in no. divisible by 2
  which are subtracted from 100
  so 100-50= 50
  and so on we can do in such a way
  
• 8 years agoHelpfull: Yes(1) No(1)
• https://www.quora.com/Is-there-some-shortcut-way-of-finding-the-count-for-the-following-question-instead-of-crunching-numbers-leading-to-some-human-mistake-Find-the-number-of-numbers-between-100-to-400-which-are-divisible-by-either-2-3-5-and-7
• 7 years agoHelpfull: Yes(1) No(0)
• Guys I know most of you are confused with the answer. So here is a simple c++ program which is giving answer as 232.
  #include
  using namespace std;
  
  int main(){
  int cnt=0;
  for(int i=100;i
• 7 years agoHelpfull: Yes(1) No(0)
• LCM of 2,3,5,7=210.
  So 210k is divisible by 2,3,5,7.
  So only one no is there.....
  
• 9 years agoHelpfull: Yes(0) No(20)
• Let S(n)= number of divisors of n between 100 and 400 (both included)
  S(n)=(300/n) +1 if 300 is divisible by n
  S(n)=floor(300/n) if 300 is not divisible by n
  
  Using Inclusion-Exclusion Principle:
  S(2|3|5|7)=S(2)+S(3)+S(5)+S(7)
  -S(2&3)-S(2&5)-S(2&7)
  -S(3&5)-S(3&7)
  -S(5&7)
  -S(2&3&5)-S(2&3&7)-S(3&5&7)
  -S(2&3&5&7)
  =151+101+61+42
  -51-31-21
  -21-14
  -8
  -11-7-2
  -1
  = 188
• 9 years agoHelpfull: Yes(0) No(3)
• ans is 210
  bcz 210 is d only 1 number bwn 100 and 400 divisible by 2,3,5,7
  divisiblity by 2 d last digit 2,4,6,8,0
  5 d last digit 5,0
  
• 9 years agoHelpfull: Yes(0) No(6)
• answer is 230
• 8 years agoHelpfull: Yes(0) No(0)
• 149+50+30+16=245
• 8 years agoHelpfull: Yes(0) No(0)
• Correct answer is 233👍but can someone give a detailed explanation...i tried calculating through formulae but failed..so calculated the answer manually😅
• 7 years agoHelpfull: Yes(0) No(2)
• So here is a simple solution to this question...firstly write down the numbers from 105 that are divisible by 7...105,112,119.....391
  Calculate the number of numbers only divisible by 7 and not by 2,3 and 5.
  119,133,161,203,217,259,287,301,329,343,371,391. 12 number
  Next calculate the numbers divisible by 2 using formula ((last number - first number)/2) +1= (400-100)/2 + 1 = 151
  Similarly Calculate the numbers divisible by 3 = 50
  Now from 105 write down all the numbers with gap of 10. 105,115,125,135,.....395.
  Calculate the numbers divisible by only 5 and not by 3
  Once you start calculating you will encounter a pattern. 105(divisible by 3) 115 & 125(only divisibly by 5 and not by 3) 135(divisible by 3) we need to count the numbers divisible by only 5 i.e the middle two numbers.
  115,125,145,155,175,185,205,215,235,245,265,275,295,305,325,335,355,365,375,385.
  Count comes to 20.
  Now add all of them
  151+50+20+12 = 233.
  Cheers!!!
• 7 years agoHelpfull: Yes(0) No(2)
• nly one.
  
  All the numbers 2,3,5,7 are prime numbers.
  
  In order to get the number which is divisible by all those primes ,it must be a multiple of 2*3*5*7= 210
  
  So it must be a multiple of 210.
  
  So there is only one number 210 between 100 and 400
• 7 years agoHelpfull: Yes(0) No(0)
• This can be solve using Inclusion and Exclusion principle:
  In its simplest form it states that if you have two finite sets,
  S1,S2,S1,S2, then their sizes are related by
  
  ∣S1∪S2∣=∣S1∣+∣S2∣−∣S1∩S2∣
  
  ∣S1∪S2∣=∣S1∣+∣S2∣−∣S1∩S2∣
  
  This is clear once one recognizes that the terms ∣S1∣+∣S2∣∣S1∣+∣S2∣ on the right count the elements in the intersection twice, so we correct the count by subtracting the elements in the intersection to get the total number of elements in either of S1S1 or S2.S2.
  This holds for more than two sets. In particular, for four sets we'll have
  ∣S1∪S2∪S3∪S4∣=∣S1∣+∣S2∣+∣S3∣+∣S4∣
  −∣S1∩S2∣−∣S1∩S3∣−⋯−∣S3∩S4∣+∣S1∩S2∩S3∣+⋯+∣S2∩S3∩S4∣−∣S1∩S2∩S3∩S4∣
• 6 years agoHelpfull: Yes(0) No(0)