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Need help on how to solve it quickly

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Need help on how to solve it quickly

by garuhape » Wed Mar 16, 2011 3:41 pm
Hey there, I'm not sure how to solve this one quickly:

How many numbers between 200 and 3600 inclusive are divisible by 4, 5 and 6?

We search for numbers which are not divisible by 2,3,4,5. So how can we quickly figure out which numbers are either prime numbers or dividable by 7?

Thx
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by srcc25anu » Wed Mar 16, 2011 4:26 pm
numbers divisible by 4: [(3600-200) / 4] +1 = 851

numbers divisible by 5: [(3600-200) / 5] +1 = 681

numbers divisible by 6: [(3600-200) / 6] +1 = 567

is this what you were looking for?
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by srcc25anu » Wed Mar 16, 2011 4:32 pm
i am afraid i got the wrong interpretation. i think the Q is asking for how many nos are divisible by all 3 (4,5 and 6)
in that case, LCM of 4,5 and 6 would be 60 so we find number of numbers between 200 and 3600 divisible by 60 or (3600-200) / 60 and add 1 to the solution
i.e. 57
what's the answer by the way?
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by Night reader » Wed Mar 16, 2011 4:46 pm
hi there, you almost got it right - the LCM (lowest common multiplier) should include 2^2,3,5 (the primes) OR the numbers should be Either divisible by 60 OR Not divisible by 60. So, the question turns to be 'How many numbers between 200 and 3,600 inclusive, which can be divided by 60?'

As you see 200 cannot be divided by 60, because 200=2^3 * 5^2. We need the numbers which are divisible by 60 to include at least all of our primes 2^2,3 and 5. In 200, one 2 is extra, 3 is missing and one 5 is extra. So if we multiply 200 by 3 we get 600 which is within the scope.

Simply putting, we neeed to find 3,600 includes how many 60 number sets? 3,600/60=60 hence we have 60 sixties ... And out of this 60 sixties, we can subtract the first 3 sixties because the first 3 sixties will make 180 and we need the numbers starting 200. So our answer would be (60-3)=57

IOM 57 numbers
garuhape wrote:Hey there, I'm not sure how to solve this one quickly:

How many numbers between 200 and 3600 inclusive are divisible by 4, 5 and 6?

We search for numbers which are not divisible by 2,3,4,5. So how can we quickly figure out which numbers are either prime numbers or dividable by 7?

Thx
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by rohu27 » Wed Mar 16, 2011 5:31 pm
thanks night reader. tht was great.
wht if the Q asked the list of numbers which are divisible byr 4 or 5 or 6? we would need more manual calcuation then or is thr a way around?
Night reader wrote:hi there, you almost got it right - the LCM (lowest common multiplier) should include 2^2,3,5 (the primes) OR the numbers should be Either divisible by 60 OR Not divisible by 60. So, the question turns to be 'How many numbers between 200 and 3,600 inclusive, which can be divided by 60?'

As you see 200 cannot be divided by 60, because 200=2^3 * 5^2. We need the numbers which are divisible by 60 to include at least all of our primes 2^2,3 and 5. In 200, one 2 is extra, 3 is missing and one 5 is extra. So if we multiply 200 by 3 we get 600 which is within the scope.

Simply putting, we neeed to find 3,600 includes how many 60 number sets? 3,600/60=60 hence we have 60 sixties ... And out of this 60 sixties, we can subtract the first 3 sixties because the first 3 sixties will make 180 and we need the numbers starting 200. So our answer would be (60-3)=57

IOM 57 numbers
garuhape wrote:Hey there, I'm not sure how to solve this one quickly:

How many numbers between 200 and 3600 inclusive are divisible by 4, 5 and 6?

We search for numbers which are not divisible by 2,3,4,5. So how can we quickly figure out which numbers are either prime numbers or dividable by 7?

Thx
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by OneTwoThreeFour » Wed Mar 16, 2011 6:29 pm
You don't need to do it manually, but its def gonna take a bit more time. First find all the numbers that are divisible by 4, 5, or 6 between 200 and 3600. Now comes the tricky part: Since some of the numbers overlap, (IE: 240 is divisible by 4,5,6 so its gonna appear three times in the total sum of all the numbers that are divisible by 4, 5, or 6) you only need to keep each unique number that are some multiples of 4,5, or 6. (IE: You only need to keep one 240, not three of them.) Thus subtract all the excessive multiples of (4)(5), (4)(6), (6)(5), and (4)(5)(6).
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