Latest TCS Aptitude Question SOLUTION: 22. How many even integers n, where 100  n  200, are divisible neither by seven nor by nine?

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

Asked In TCS Bishal Pandey (10 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

manofmayhem  (10 years ago)

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

Is this solution Helpfull? Yes (24) | No (8)  

rk  (10 years ago)

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)

Is this solution Helpfull? Yes (17) | No (7)  

VISHAKHA BHATTACHARJEE  (10 years ago)

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

Is this solution Helpfull? Yes (8) | No (2)  

Anay Das  (10 years ago)

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

Is this solution Helpfull? Yes (7) | No (7)  

mritunjay kumar  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No (6)  

Naba Paul  (10 years ago)

@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

Is this solution Helpfull? Yes (2) | No (4)  

VISHAKHA BHATTACHARJEE  (10 years ago)

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

Is this solution Helpfull? Yes (1) | No (3)  

chandan kumar  (10 years ago)

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

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arijit nandy  (8 years ago)

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

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Garima Pathak  (8 years ago)

100

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Rahul Meena  (7 years ago)

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.)

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AKARSH SRIVASTAVA  (6 years ago)

We can find this by removing numbers divisible by either 7 or 9 or both from all possible even numbers, where 100

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