How many integers between 100 and 150, inclusive, cannot be
evenly divided by 3 nor 5?
Answers were Sorted based on User's Feedback
Answer / prabhu
number me
integers
between then is
=51
the number of
integers divided
by 3 of the
numbers
=17
numere of
integers divided
by 5 of
there number
=11
number OF
integers divided
by both =4
so that
17+11_4=24
and 51-24=27
| Is This Answer Correct ? | 45 Yes | 8 No |
Answer / ramabramham
Number of integers that divide 3:
the range is 100-150
Relevant to this case, we take 102 - 150 (since 102 is the
first to div 3)
102 = 34*3
150= 50*3, so we have 50-34+1 = 17 multiples of 3
For multiples of 5,
100=5*20
150=5*30
30-20+1 =11
Now we have a total of 27 integers, but we double counted
the ones that divide BOTH 3 AND 5, ie 15.
105 is the first to divide 15.
105=15*7
150=15*10
10-7+1 = 4 integers
So our total is 17+11-4 = 24 integers that can be divided
by either 3 or 5 or both.
51 integers - 24 integers = 27 that cannot be evenly divided
| Is This Answer Correct ? | 25 Yes | 0 No |
Answer / nupur
its 27 because the no divisible by three is 17 and that by
5 is 11 common nos for both will be 4 hence our ans will be
neither for 5 nor for three therfore it is 49-17-11-4=27
| Is This Answer Correct ? | 15 Yes | 1 No |
Answer / kondayya
ans:27
there are 24 numbers which are divisible by
both 3 and 5.
51-24=27
| Is This Answer Correct ? | 7 Yes | 1 No |
Answer / tamilselvi
ans=4
no divisible by 3:17
A =
{102,105,108,111,114,117,120,123,126,129,132,135,138,141,144
,147,150}
no divisible by 5: 11
B= {100,105,110,115,120,125,130,135,140,145,150}
3 nor 5 means ^(AvB)
So, AvB = 25
^(AvB) = 4
now strike out 105 120 135 150,as it is present in
both.remaining totally 24 nos are present,which cannot be
evenly divided 3 or 5.(means wen divisible by 3 it is not
by 5 n vice versa)
| Is This Answer Correct ? | 3 Yes | 6 No |
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