Fastest way?

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Tazbo

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Hi,

I get these kind of questions all the time is there a quick way I can work them out in an exam?Thank you.

Samantha has some marbles
The marbles can be divided exactly into piles of 3,4 or 5

What is the smallest number of marbles Samantha could have?
 
Also please help with this


There are 15 yellow balls, 10 blue balls and 8 green balls mixed up in a bag.

Paul takes the balls from the bag one by one at random.

He wants to take out at least 2 balls of each colour.
What is the least number of balls that Paul needs to take from the bag to guarantee doing this?
 
The second problem: The largest number of balls of one color is the 15 yellow balls. It is possible that all 10 blue balls and 8 green balls could all be taken out before any yellow balls. So At least 10+ 8+ 2= 20 balls must be take out in order to be certain of getting at lest two balls of each color.
 
HallsofIvy said:
The second problem: The largest number of balls of one color is the 15 yellow balls. It is possible that all 10 blue balls and 8 green balls could all be taken out before any yellow balls. So At least 10+ 8+ 2= 20 balls must be take out in order to be certain of getting at lest two balls of each color.
Click to expand...
Isn't the worst case getting all yellow and blue first?
 
HallsofIvy said:
The second problem: The largest number of balls of one color is the 15 yellow balls. It is possible that all 10 blue balls and 8 green balls could all be taken out before any yellow balls. So At least 10+ 8+ 2= 20 balls must be take out in order to be certain of getting at lest two balls of each color.
Click to expand...
Professor Halls,
How about picking all 15 yellow balls and 10 blue balls before getting the 2nd 8 green ball. This case requires that you must pick 27 balls.

This is exactly what Lev888 said.
 
Tazbo said:
I get these kind of questions all the time is there a quick way I can work them out in an exam?Thank you.

Samantha has some marbles
The marbles can be divided exactly into piles of 3,4 or 5

What is the smallest number of marbles Samantha could have?
Click to expand...
What methods do you know for finding a least common multiple? You may or may not know the quickest way (which varies according to how large the numbers are).

By the way, you can already see why we ask you to put only one question in a thread ...
 
Tazbo said:
Hi,

I get these kind of questions all the time is there a quick way I can work them out in an exam?Thank you.

Samantha has some marbles
The marbles can be divided exactly into piles of 3,4 or 5

What is the smallest number of marbles Samantha could have?
Click to expand...
Let's try something more interesting: 4, 6, 15
Prime factors for each:
4 = 2*2
6 = 2*3
15 = 3*5

The number we are looking for is divisible by 4, therefore, it has to have 2*2 as prime factors. N = 2*2...
It's divisible by 6, so it has to have 2 and 3 as prime factors. But we know there is a 2 already, so we need to add only 3: N = 2*2*3...
It's divisible by 15, so it has to have 3 and 5 as prime factors. But we know it has 3 already, so we need to add only 5: N = 2*2*3*5 = 60.
 
Answer was 27.

Worst case scenario is he takes all 15 yellow, then all 10 blue, and finally 2 green. Chances are he'd need to take out fewer than 27, but that doesn't guarantee 2 of each colour
 
Correct.

I like to approach these problems by imagining that as I take marbles randomly, I have an enemy manipulating things behind the scenes, trying to keep me from achieving my goal, in this case getting two of each, for as long as possible. That would involve giving me all 15 yellow and all 10 blue before I get the second green. That is, hold the smallest set for last. The enemy's strategy is easier to figure out than my own random expectations ...
 
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