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07 Jul 2016, 05:16
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Q--how many prime factors does r have if r/10 and r/15 is an integer?
10-> 2*5
15-> 3*5
does r have to be at least 2*3*5*5?
10-> 2*5
15-> 3*5
does r have to be at least 2*3*5*5?
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07 Jul 2016, 12:32
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r only needs one 5 in it's PF.
I consider the numbers one at a time and keep a running track of the minimum that I need:
In order for r/10 to be an int, r needs at least one 2 and one 5. So now we have 2*5.
In order for r/10 to be an int, r needs at least one 3 and one 5. We already have a 5, so this just adds a 3; now we have 2*3*5.
so 2*3*5 is the absolute minimum that must be in r's PF. You can confirm that his works; 2*3*5 = 30. 30 is a mult of both 10 and 15 and there is no smaller number that is a mult of both 10 and 15.
Hope this helps.
I consider the numbers one at a time and keep a running track of the minimum that I need:
In order for r/10 to be an int, r needs at least one 2 and one 5. So now we have 2*5.
In order for r/10 to be an int, r needs at least one 3 and one 5. We already have a 5, so this just adds a 3; now we have 2*3*5.
so 2*3*5 is the absolute minimum that must be in r's PF. You can confirm that his works; 2*3*5 = 30. 30 is a mult of both 10 and 15 and there is no smaller number that is a mult of both 10 and 15.
Hope this helps.
07 Jul 2016, 21:38
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ha15
You are thinking about successive division.
Say, r = 2*3*5*5
r/10 = 3*5 is an integer.
Now 3*5/15 = 1 is an integer.
This is when r must be at least 2*3*5*5 (you still can't say how many prime factors is has since 2*3*5*5*7*11 etc is also divisible by both 10 and 15)
But here, you are not given that r is divided successively.
r/10 = 2*3*5/10 = 3 (an integer)
r/15 = 2*3*5/15 = 2 (an integer)
So it is enough to have just a single 5. Only in case of successive division (when the 5 is canceled off in the first step) do you need two 5s. Since it is not mentioned that you need to divide success
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
