Latest Elitmus Aptitude Question SOLUTION: Find the number of number between 100 to 400 which are divisible by either 2 ,3 ,5 and 7

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(#M40088170) ELITMUS Numerical Ability question number system Keep an EYE

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

Asked In Elitmus sarojani (9 years ago)

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Abhishek singh  (9 years ago)

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

Is this solution Helpfull? Yes (15) | No (5)  

CHINMOY MANDAL  (9 years ago)

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.

Is this solution Helpfull? Yes (10) | No (14)  

paras  (8 years ago)

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

Is this solution Helpfull? Yes (10) | No   

rahul  (7 years ago)

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.

Is this solution Helpfull? Yes (9) | No   

ankit   (8 years ago)

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.

Is this solution Helpfull? Yes (2) | No (2)  

jyoti kumari choudhary  (9 years ago)

LET of 2 3 5 7= 210
Ans is 210 numbers

Is this solution Helpfull? Yes (1) | No (16)  

sourav  (9 years ago)

1 as take lcm of 2,3,5,7 that is 210

Is this solution Helpfull? Yes (1) | No (8)  

Atik  (9 years ago)

ans is 231... i manually calculate the no.

Is this solution Helpfull? Yes (1) | No (2)  

Pawan Kumar  (8 years ago)

the lcm of 2,3,5,7 is 210 so only one is answer

Is this solution Helpfull? Yes (1) | No (6)  

vinay kumar  (8 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

Soumya Priyadarsini  (7 years ago)

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

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Sumanth Sunny  (6 years ago)

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

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Saroj kumar  (9 years ago)

LCM of 2,3,5,7=210.
So 210k is divisible by 2,3,5,7.
So only one no is there.....

Is this solution Helpfull? Yes | No (20)  

NITHISH  (8 years ago)

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

Is this solution Helpfull? Yes | No (3)  

sandhya ambatii  (8 years ago)

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

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Abhijeet Jain  (7 years ago)

answer is 230

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mahir madan  (7 years ago)

149+50+30+16=245

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Ajinkya bhosale  (7 years ago)

Correct answer is 233👍but can someone give a detailed explanation...i tried calculating through formulae but failed..so calculated the answer manually😅

Is this solution Helpfull? Yes | No (2)  

Ajinkya bhosale  (7 years ago)

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!!!

Is this solution Helpfull? Yes | No (2)  

Madhu  (6 years ago)

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

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Gourav Khatter  (6 years ago)

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∣

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