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Re: How many numbers between 1 and 250, excluding, are divisible [#permalink]
[quote="GreNow"]need to multiply the no of multiples of either 5 or 7 but not both by 2 as they've been counted twice.

no of multiples of either 5 or 7 but not both 49+35-7(2)=70

answer B

Why 7 is multiplied by 2? Explain plz
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Re: How many numbers between 1 and 250, excluding, are divisible [#permalink]
can there be any shortcut for such type of problems
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Re: How many numbers between 1 and 250, excluding, are divisible [#permalink]
3
GeminiHeat wrote:
How many numbers between 1 and 250, excluding, are divisible by either 5 or 7 but not both?

(A) 69
(B) 70
(C) 71
(D) 77
(E) 78



Number of integers divisible by either 5 or 7 but not both = (number of integers divisible by 5) + (number of integers divisible by 7) - 2(number of integers divisible by 5 AND 7)
Aside: In a moment, I'll explain why we're multiplying the last value by 2

Number of integers divisible by 5
The numbers are: 5, 10, 15, ....., 240 and 245
We can rewrite these numbers as follows: (1)(5), (2)(5), (3)(5), ...(48)(5) and (49)(5)
Number of integers divisible by 5 = 49

Number of integers divisible by 7
The numbers are: 7, 14, 21, .....,238, 245
We can rewrite these numbers as follows: (1)(7), (2)(7), (3)(7), ...(34)(7) and (35)(7)
Number of integers divisible by 7 = 35

NOTICE THAT BOTH LISTS INCLUDE 245
It's also true that both lists include 35, 70, 105 and all other multiples of 35
So, if we want to exclude all numbers that are divisible by 5 AND 7, we must keep in mind that all multiples of 35 have been counted TWICE

Number of integers divisible by 5 AND 7
But this is the same as finding the number of integers divisible by 35
They are: 35, 70, 105, ....210 and 245
We can rewrite these numbers as follows: (1)(35), (2)(35), (3)(35), ...(6)(35) and (7)(35)
Number of integers divisible by 5 AND 7 = 7


Number of integers divisible by either 5 or 7 but not both = 49 + 35 - 2(7)
= 70

Answer: B

Cheers,
Brent
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Re: How many numbers between 1 and 250, excluding, are divisible [#permalink]
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