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Logical Reasoning
Number Series
22. How many even integers n, where 100 n 200, are divisible neither by seven nor by nine?
Read Solution (Total 12)
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- 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
- 11 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)
- 11 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 - 11 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 - 11 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 - 11 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
- 11 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 - 11 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 - 11 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 - 9 years agoHelpfull: Yes(0) No(1)
- 100
- 9 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.) - 8 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)
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