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Re: A pub owner has three different kinds of beer of quantities [#permalink]
:good
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
:good
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
Why isn't it three?

The question does not state that the canisters must be filled to the rim.
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
Stoolfi is doing some serious CR - Weaken the question lately :wink:
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
:lol:
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
82+123+205 = 410

this can be factorised as 41*10
So the minimum number of cans required to completely fill each can without mixing = 10

Again largest common factor among them is 41. The problem can be solved this way as well
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
I see. But I still think it's better by breaking it down to factors. It's easier to see the 41 stand out. But for an accustomed eye, your way is sure easier
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
I didn't like this problem. We could have a canister of 82 liters, and in that case the answer would be 5
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Re: A pub owner has three different kinds of beer of quantities [#permalink]
It cannot be 82 because it is not a common factor to 123 and 205...



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