Latest Elitmus Aptitude Question SOLUTION: let S be a set of postive integers such that every element n of S satisfies the condition 1-1000<=n<=1300 2-every digit in n is odd then how many element of S are divisible b

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(#M40081843) ELITMUS Numerical Ability question number system Keep an EYE

let S be a set of postive integers such that every element n of S satisfies the condition
1-1000<=n<=1300
2-every digit in n is odd
then how many element of S are divisible by 3 ???
a-9
b-15

Asked In Elitmus azad yadav (9 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

Manju Prakash R  (9 years ago)

a-9
elements of S exists between 1110 and 1199.
Elements are 1113, 1119, 1131, 1137, 1155, 1173, 1179 , 1191, 1197

Is this solution Helpfull? Yes (27) | No (9)  

Mithun Kumar Kesari  (9 years ago)

possible combinations are
(1,1,1,3)= 2 ways
(1,1,1,9)= 2 ways
(1,1,3,7)= 2 ways
(1,1,5,5)=1 way
(1,1,7,9)= 2 way

so total 9 numbers are possibles between 1000 and 1300 which are divisible by 3.

Is this solution Helpfull? Yes (19) | No (3)  

Moksh Juneja  (9 years ago)

n is between 1000-1300 and should be odd and divisible by 3..
so first no. is 1005
last no. is 1299

after 1002 is even 1005 is odd and again 1008 is even
so next number is 1011.. so after difference of 6 again we get odd and divisible by 3.

so now using A.P

1299=1005 + (n-1)6
we get n=50

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

Abhishek kumar Singh  (9 years ago)

as per the given question.
n is between 1000 and 1300 and also each digit is odd no. so that no. is as follows
1111,1113,1115,1117,1119
1131,1133...................,1139
.
.
.
.
.
1191....................................1199

we found the value of n now you have to see which no. is divisible by 3


so that ans=9

Is this solution Helpfull? Yes (10) | No (2)  

Ehtesham  (9 years ago)

i think given options for answer is incorrect.
Answer would be 50.
possible elements of n={1005,1011,1017,...1299}
so here we can see that the diff. between two number is 6.so total number divisible by 3
=((1299-1005)/2) + 1
=50

Is this solution Helpfull? Yes (6) | No (7)  

akshay kumar  (8 years ago)

Manju why u have not started with 1005 ,1017 this number are also divisible by 3 and all digits are odd.

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

LAKHAN SINGH  (9 years ago)

l=1299
a=1002
l=a+(n-1)d
so that n=100

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

Life  (9 years ago)

n>=1000 so first such no. can be 1005 since it is odd and divisible by 3
n

Is this solution Helpfull? Yes (1) | No (4)  

prashant shekhar singfh  (8 years ago)

given options are incorrect
1st number that satisfies the condition is 1005 then 1011,1017....(common difference is 6)
last number will be 1299 (1005,1011,1017............,1293,1299) this is the series
using AP(a=1005, d=6) 1299=1005+(n-1)6
answer is n=50

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

shilpa  (7 years ago)

44 becoz 1000 to 1100 ->14
1100 to 1200->16
1200 to 1300->14
14+14+16

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Digesh Singh Bisht  (1 year ago)

Those who are saying 9 are wrong.
Here is the solution
series will be (1005,1011,1017,1023........ 1239,1299) They all are the odd no. and divisible by 3
So
a=1005 d=6 last digit(l)=1299
l=a + (n-1)d
1299 = 1005 +(n-1) 6..solve this
50 = n Answer

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