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24 May 2022, 01:31
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:50) correct
28%
(02:24)
wrong
based on 67
sessions
History
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Time
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Not Attempted Yet
How many integers from 1 to 125, inclusive, are divisible by 3 or 7 or both?
(A) 51
(B) 53
(C) 54
(D) 57
(E) 59
Are You Up For the Challenge: 700 Level Questions
(A) 51
(B) 53
(C) 54
(D) 57
(E) 59
Are You Up For the Challenge: 700 Level Questions
24 May 2022, 01:42
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Divisible by 3:
[125][/3] -> 41
Divisible by 7:
[125][/7] -> 17
Divisible by 21:
[125][/21] -> 5
Now the no.s divisible by both 3 and 7 are counted two times: in the list of no.s divisible by 3 and also in the list of no.s divisible by 7.
Therefore, we get 41 + 17 - 5 = 53
Posted from my mobile device
[125][/3] -> 41
Divisible by 7:
[125][/7] -> 17
Divisible by 21:
[125][/21] -> 5
Now the no.s divisible by both 3 and 7 are counted two times: in the list of no.s divisible by 3 and also in the list of no.s divisible by 7.
Therefore, we get 41 + 17 - 5 = 53
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24 May 2022, 01:48
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The concept is straightforward; we just need to be super careful to ensure that we get the point.
From 1 to 125,
41 integers are divisible by 3.
17 integers are divisible by 7.
5 integers are divisible by 21.
Total is 41 + 17 - 5 = 53.
The answer is thus (B)
From 1 to 125,
41 integers are divisible by 3.
17 integers are divisible by 7.
5 integers are divisible by 21.
Total is 41 + 17 - 5 = 53.
The answer is thus (B)
