# cambridge interview for maths

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#41

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didnt know that tan (x/2) will work as well.. how did u realise this?

**integral_neo**)didnt know that tan (x/2) will work as well.. how did u realise this?

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#42

(Original post by

It's a pretty common method for integrating with a number and a linear (i.e. no powers) trig function in the denominator if it's not obvious how to do it.

**theone**)It's a pretty common method for integrating with a number and a linear (i.e. no powers) trig function in the denominator if it's not obvious how to do it.

do this one:

r = 3cox(2x)

find the length between the two parallel (to the initial line) tangents to the graph

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#43

(Original post by

For the first one, write the consecutive numbers

n(n+1)(n+2)(n+3), then if n is odd, n+1, and n+3 are multiples of 2. Either n+1 or n+3 is a multiple of 4. n or n+2 must be a multiple of 3.

n - even. Same arguements.

2) A great trick i read. The highest power of a prime into a factorial is found as follows.

Sum from i=1 to infinity of [20/2^i]

Where [x] is the greatest integer function.

Lol, except it doesnt seem to work.

**JamesF**)For the first one, write the consecutive numbers

n(n+1)(n+2)(n+3), then if n is odd, n+1, and n+3 are multiples of 2. Either n+1 or n+3 is a multiple of 4. n or n+2 must be a multiple of 3.

n - even. Same arguements.

2) A great trick i read. The highest power of a prime into a factorial is found as follows.

Sum from i=1 to infinity of [20/2^i]

Where [x] is the greatest integer function.

Lol, except it doesnt seem to work.

wot year u in james?

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#44

(Original post by

Are you doing Edexcel maths and F maths? what modules have u taken and what UMS did u get for each and which ones u will taking in June? r u gonna do extension paper?

do this one:

r = 3cox(2x)

find the length between the two parallel (to the initial line) tangents to the graph

**integral_neo**)Are you doing Edexcel maths and F maths? what modules have u taken and what UMS did u get for each and which ones u will taking in June? r u gonna do extension paper?

do this one:

r = 3cox(2x)

find the length between the two parallel (to the initial line) tangents to the graph

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#45

(Original post by

I'm not going to go through this again I remember it from my P4 exam and took me half a side and I don't have time to regurgitate. I'll pm my marks/modules.

**theone**)I'm not going to go through this again I remember it from my P4 exam and took me half a side and I don't have time to regurgitate. I'll pm my marks/modules.

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#46

No, i is not sqrt(-1). Read what i said about the sum. Using it for this question we get

k = (sum from i=1 to infinity) [20/2^i]

2^5 > 20 so we only need to consider the cases, when i is 1,2,3,4.

[20/2] = 10

[20/4] = 5

[20/8] = 2

[20/16] = 1

Sum these, 10+5+2+1 = 18.

Im in Lower 6th.

k = (sum from i=1 to infinity) [20/2^i]

2^5 > 20 so we only need to consider the cases, when i is 1,2,3,4.

[20/2] = 10

[20/4] = 5

[20/8] = 2

[20/16] = 1

Sum these, 10+5+2+1 = 18.

Im in Lower 6th.

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#47

For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4

[20/25] = 0

So the answer is 4.

20/5 = 4

[20/25] = 0

So the answer is 4.

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#48

(Original post by

For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4

[20/25] = 0

So the answer is 4.

**JamesF**)For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4

[20/25] = 0

So the answer is 4.

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#49

**JamesF**)

For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4

[20/25] = 0

So the answer is 4.

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#52

(Original post by

It was in some university lecture notes on elementary number theory.

**JamesF**)It was in some university lecture notes on elementary number theory.

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#53

No I didnt do SMC.

I cant remember where i found them, but google will throw up loads of stuff if you search.

I cant remember where i found them, but google will throw up loads of stuff if you search.

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