Latest Elitmus Aptitude Question SOLUTION: how many no. are between 1100 - 1300 which are divisible by 3 but condition is that all 4-digits of no. should be odd for example-1131

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(#M40019901) ELITMUS question number system Keep an EYE

how many no. are between 1100 - 1300
which are divisible by 3
but condition is that all 4-digits of no. should be odd
for example-1131

Asked In Elitmus Sahil (10 years ago)

Unsolved Read Solution (15)

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rahul  (10 years ago)

1300-1100=200 now divide it by 2 we have 100 even number hence 100 odd no's now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197 hence
9 no's

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Rupesh Kumar  (10 years ago)

ans:34
1st no divisible by 3 in b/w 1100-1300 is 1101.
so if we see the pattern which is divisible by 3 and shoulb be odd like this
1101,1107,1113,1119,1125,1131..............................1299.
d=6
an=1299
so now,1299=1101+(n-1)6
=>n=34

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Rupesh Kumar  (10 years ago)

Amit bhai, total no are-9.
1113,1119,1131,1137,1155,1173,1179,1191,1197.

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VIKAS SINGH  (10 years ago)

so according to given condn we hv to find only d no. bw 1100 to 1200 because after that all no. have 2 which is even 1201 to 1299. so nos which setesfies above condn r 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1195,.
so there r total 12 no.s.
so ans will be 12.

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Amit Agarwal  (10 years ago)

Sorry the answer will be 13.
I missed out d 2nd coditon in my 1st answer(All four digits shall be odd).

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Sahil  (10 years ago)

Meghna baby 0 odd ni h...so answer is 9

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VIKAS SINGH  (10 years ago)

sorry ans will be 13 i missed the no. 1197.

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VIKAS SINGH  (10 years ago)

sorry ans will be 12 nd no.s will be 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1197

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meghana  (10 years ago)

answer is 13
no's are 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1191,1197
this satisfy's both the conditions

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Amit Agarwal  (10 years ago)

67 no.s are divisible

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Rupesh Kumar  (10 years ago)

Amit please explain your ans.I know my ans is wrong.

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kratika jain  (10 years ago)

34 nos
11_ _ last digit 1 (2nd pos 0 3 6 9)= 4 sum 3
last digit 3 (2nd pos 1 4 7)= 3 sum 5
last digit 5 (2nd pos 2 5 8)= 3 sum 7
last digit 7 (2nd pos 0 3 6 9)= 4 sum 9
last digit 9 (2nd pos 1 4 7)=3 sum 11 tot 17
12_ _ last digit 1 (2nd pos 2 5 8)= 3 sum 4
last digit 3 (2nd pos 0 3 6 9 )= 4 sum 6
last digit 5 (2nd pos 1 4 7)= 3 sum 8
last digit 7 (2nd pos 2 5 8)= 3 sum 10 tot 13
last digit 9 (2nd pos 0 3 6 9)=4 sum 11 tot 17

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swati ahuja  (10 years ago)

considering the second condition,numbers between 1200 to 1300 are of no use as they contain even digits.
so between 1100 and 1199....no of numbers divisible by 3 are=30 (since 3 nos in between 1111 & 1120..similarly in each series of 10 nos,there will be 3 nos divisible by 3...so,3*9=27......also 1110,1140,1170 are divi by 3...hence,27+3=30).
now,the 3 nos in each series with even nos is eliminated i.e 1121-1129,1141-1149,1161-1169,1181-1189. so,4*3=12 nos are eliminated.
also 1110,1140,170 i.e 3 nos are eliminated.
hence,30-12-3=15.
now 1116,1134,1152,1158,1176,1194 contain even so,6 nos are eliminated.
total=15-6=9.

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prayas goel  (10 years ago)

I think is must be 34
Between 1100-1300
no. divisible by 3 and also should be odd..
i.e. starting from 1101 ....next will be 1107..
So series will be 1101, 1107, 1113...upto 1299
total terms will be 34..agree with Rupesh

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niranjan kumar  (10 years ago)

ans should be 19

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