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5*1200=6000, so 5*n, where n is any natural number between 1 and 1200, is less than 6000 and a multiple of 5.

(of course, if counting numbers that are multiples of 2 or 3 or 5, you have to be careful not to "doublecount" numbers that are both multiples of 5 and 2, or 5 and 3, etc.)

(of course, if counting numbers that are multiples of 2 or 3 or 5, you have to be careful not to "doublecount" numbers that are both multiples of 5 and 2, or 5 and 3, etc.)

for the set of all integers 1 to 6000 inclusive, there are obviously 6000/2=3000 elements that are divisible by 2. 6000/3=2000 elements that are divisible by 3 and 6000/5=1200 that are divisible by 5. The sum of these is 3000+2000+1200=6200. You must now subtract the numbers which we have counted multiple times. e.g. number 6 will be counted twice as it is divisible by 3 and 2. all the numbers divisible by both 2 and 3 will be multiples of 6, so there are 6000/6=1000 of these such numbers. all the numbers divisible by 2 and 5 will be multiples of 10, so there are 6000/10=600 of these such numbers. all the numbers divisible by 3 and 5 are multiples of 15, so there are 6000/15=400 of these such numbers. However the numbers divisible by all of 2,3 and 5 have been removed 3 times so must be added again. There are 6000/(2*3*5)=6000/30=200 numbers divisible by 2,3 and 5. Therefor the number of multiples of 2,3 or 5, in the integers 1-6000 is 3000+2000+1200-1000-600-400+200=4400. therefore, the number of integers that are not multiples is 6000-4400=1600.

OMFG

I GOT THAT

W00T

OK

HERE IS MY METHOD

I found the number of factors which are multiples of 2 3 or 5 under 6000 and took that from 6000

there are 3000 multiples of 2 under 6000

ok so you have 3000 so far

then there are 1200 multiples of 5 under 6000 but only 1 in 2 of these are odd so you have 600 distinct multiples of 5 under 6000

then for 3 for every 5 multiples of 3 there are only 2 distinct ones which are not a multiple of 2 or 5

so 6000/5*2=800

800+3000+600=4400 6000-4400=1600

WHY MUST PEOPLE POST INFRONT OF ME NOOOOOOO

I GOT THAT

W00T

OK

HERE IS MY METHOD

I found the number of factors which are multiples of 2 3 or 5 under 6000 and took that from 6000

there are 3000 multiples of 2 under 6000

ok so you have 3000 so far

then there are 1200 multiples of 5 under 6000 but only 1 in 2 of these are odd so you have 600 distinct multiples of 5 under 6000

then for 3 for every 5 multiples of 3 there are only 2 distinct ones which are not a multiple of 2 or 5

so 6000/5*2=800

800+3000+600=4400 6000-4400=1600

WHY MUST PEOPLE POST INFRONT OF ME NOOOOOOO

haha, got in there first

3000 are multiples of 2,

1200 are multiples of 5

2000 are multiples of 3,

multiples of 5 end on 0 or 5, so half off all multiples of 5 are also multiples of 2, so 3600 are multiples of 2 and/or 5

3->30 (repeat starts)

3 (none)

6(2)

9(none)

12(2)

15(5)

18(2)

21(none)

24(2)

27(none)

30(2)

so only 4 out of 30 numbers havent been counted yet, so 4/10*2000=800

3600+800=4200. So 4200 numbers which are multiples. Now do 6000-4200, giving you the solution

1200 are multiples of 5

2000 are multiples of 3,

multiples of 5 end on 0 or 5, so half off all multiples of 5 are also multiples of 2, so 3600 are multiples of 2 and/or 5

3->30 (repeat starts)

3 (none)

6(2)

9(none)

12(2)

15(5)

18(2)

21(none)

24(2)

27(none)

30(2)

so only 4 out of 30 numbers havent been counted yet, so 4/10*2000=800

3600+800=4200. So 4200 numbers which are multiples. Now do 6000-4200, giving you the solution

WHY MUST PEOPLE POST INFRONT OF ME NOOOOOOO

that solution is totally wrong

Look Bud you're 3 minutes behind me and Mr Gestapo over there.

STOP DOING THAT!

STOP DOING THAT!

To avoid crossovers, consider the most frequent multiple:

Every even number is a multiple of 2, so that is 3000.

to every 3 multiples of 2, there are 2 multiples of 3, but one is a cross over (multiple of 6) so added to the 3000 multiples of 2 there are 1000 multiples of 3.

2*5 = 10 and 3*5 = 15.

This means that to every 3 multiples of 5, one is also a multiple of 2, and the other a multiple of 3. Hence if there are 1200 multiples of 5 from 1 to 6000, the multiples that are only of 5 are 400.

So those not multiples of 2,3 and 5 is 6000 - (400 + 1000 + 3000) = 1600.

SO U HAD THE RIGHT ANSWER ALL ALONG!!!.....

....or did you???

Every even number is a multiple of 2, so that is 3000.

to every 3 multiples of 2, there are 2 multiples of 3, but one is a cross over (multiple of 6) so added to the 3000 multiples of 2 there are 1000 multiples of 3.

2*5 = 10 and 3*5 = 15.

This means that to every 3 multiples of 5, one is also a multiple of 2, and the other a multiple of 3. Hence if there are 1200 multiples of 5 from 1 to 6000, the multiples that are only of 5 are 400.

So those not multiples of 2,3 and 5 is 6000 - (400 + 1000 + 3000) = 1600.

SO U HAD THE RIGHT ANSWER ALL ALONG!!!.....

....or did you???

No he didn't thats the actual answer that he couldn't get.

ANYWAYS. For question 7 on that sheet,

http://www.mathshelper.co.uk/Oxbridge%20Questions.pdf

I get there to be N games needed to be played to find the winner and no additional games needed to be played to find the second best team. The second best team will be the one that lost in the final...

OR AM I WRONG?

ANYWAYS. For question 7 on that sheet,

http://www.mathshelper.co.uk/Oxbridge%20Questions.pdf

I get there to be N games needed to be played to find the winner and no additional games needed to be played to find the second best team. The second best team will be the one that lost in the final...

OR AM I WRONG?

Hmm, just has quick go at that question and got the answer 1600, but my method didn't really seem as 'mathsy':

The LCM of 2, 3 and 5 is 30. There are 200 30s in 6000.

Only the number of non multiples in the numbers 1-30 hence needs to be found, as the pattern will repeat 199 more times:

n/x != int => (n/x)+kx != int

where n is an integer to be tested, x is the divisor (2, 3 or 5 in this case) and k is any integer.

The number of non multiples in the numbers 1-30 can be gotten just by going through them (8), and then the number of non multiples in 6000 can be gotten by multiplying this number by 200.

It'd probably need more explanation, but it got the right answer

The LCM of 2, 3 and 5 is 30. There are 200 30s in 6000.

Only the number of non multiples in the numbers 1-30 hence needs to be found, as the pattern will repeat 199 more times:

n/x != int => (n/x)+kx != int

where n is an integer to be tested, x is the divisor (2, 3 or 5 in this case) and k is any integer.

The number of non multiples in the numbers 1-30 can be gotten just by going through them (8), and then the number of non multiples in 6000 can be gotten by multiplying this number by 200.

It'd probably need more explanation, but it got the right answer

yay, now can you post the link to the site with answers for these!!Z

how did you know its 1600 then?

Incidentally, for those that care (and will read my post amongst the frenzy of other posts in this thread), the method that everyone seems to be using is called the inclusion-exclusion principle. Mattd123's method is nice too, and probably doesn't have a name.

Found it out from another web site.

For Q12 (There is a pile of 129 coins on a table, all unbiased except for one which has heads on

both sides. David chooses a coin at random and tosses it eight times. The coin comes up

heads every time. What is the probability that it will come up heads the ninth time as

well?)

i get the answer to be

Anyone agree?

For Q12 (There is a pile of 129 coins on a table, all unbiased except for one which has heads on

both sides. David chooses a coin at random and tosses it eight times. The coin comes up

heads every time. What is the probability that it will come up heads the ninth time as

well?)

i get the answer to be

Spoiler

Anyone agree?

Hmmz, I get the probability he picks an unbiased coin as 128/129 then the probability this is head 9 times is (1/2)^9

So multiplying them I get 128/(512*129)

Then I have 1/129 which is the biased head so I get 128/(512*129) + 1/129 as my probability...

Which works out to be 645/66564

which is pretty small

i think ur right

method?

So multiplying them I get 128/(512*129)

Then I have 1/129 which is the biased head so I get 128/(512*129) + 1/129 as my probability...

Which works out to be 645/66564

which is pretty small

i think ur right

method?

Just found the question. I got 1600 a different way.

2*3*5=30

So I did the question for 30 not 6000 and got 8 numbers. I did this cause I thought there would be a repeating pattern for numbers after multiples of 30, e.g. if 1 works 31 works as does 61, 91, 121 etc. So I multiplied by 200.

6000/30 = 200

200*8=1600

2*3*5=30

So I did the question for 30 not 6000 and got 8 numbers. I did this cause I thought there would be a repeating pattern for numbers after multiples of 30, e.g. if 1 works 31 works as does 61, 91, 121 etc. So I multiplied by 200.

6000/30 = 200

200*8=1600

(edited 7 years ago)

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