O-Level Additional Math

[atutor2001]

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

Using explanation from this link
http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx

You may like to read it and let me know your answer. I tried but can't get consecutive zeros

[iFruit]

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

In a factorial, every multiple of 5 will contribute one zero at the end

for example, 1x2x3x4x5 = 120, 120x6x7x8x9x10 = [something]00

In addition every multiple of 25 will contribute one extra zero

25x24, 50x48, 75 x 72, 100x99 etc.

So 120! would have 120/5 = 24 zeros contributed by multiples of 5,

and 4 extra zeros contributed by multiples of 25 (25, 50, 75, 100) with 28 zeros (24+4) at the end.

So the largest factorial with 28 consecutive zeros is 124! --> n = 124

[iFruit]

hot_chocolate:

Please help to solve the followings:



2) Find the largest no. n such that there is only one whole no. k that satisfies
8/21 < n/(n+k) < 5/13

(Note: A < C < B means that value of C is between A and B, e.g. 4 < 9 < 16)

Thank you.

8/21 < n/(n+k) < 5/13 ---> 8/21 < 1/(n+k)/n < 5/13 ---->

8/21 < 1/(1+k/n) < 5/13 ---> 1/(21/8 ) < 1/(1+k/n) < 1/(13/5) ---->

13/5 < 1 + k/n < 21/8 ----> 8/5 < k/n < 13/8 ----> 64/40 < k/n < 65/40


Now, 64/40 < k/n < 65/40 has no solutions for k = whole number, n=40

128/80 < k/n < 130/80, will have one solution for k=129, n =80

192/120 < k/n < 195/120 will have multiple solutions for k (193, 194), and n =120

so \" largest n such that there is only one whole no. k \" = 80

HTH

[atutor2001]

iFruit:

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

In a factorial, every multiple of 5 will contribute one zero at the end

for example, 1x2x3x4x5 = 120, 120x6x7x8x9x10 = [something]00

In addition every multiple of 25 will contribute one extra zero

25x24, 50x48, 75 x 72, 100x99 etc.

So 120! would have 120/5 = 24 zeros contributed by multiples of 5,

and 4 extra zeros contributed by multiples of 25 (25, 50, 75, 100) with 28 zeros (24+4) at the end.

So the largest factorial with 28 consecutive zeros is 124! --> n = 124

Hi iFruit

You are really good!

[hot_chocolate]

Hi iFruit,


The answers are correct! Thanks for the solutions.

Agree with atutor2001, you're really good.

[iFruit]

hot_chocolate:

Hi iFruit,


The answers are correct! Thanks for the solutions.

Agree with atutor2001, you're really good.

Thank you atutor2001 and hot_chocolate. You are very kind.

[emerald]

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

[iFruit]

emerald:

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

Hi emerald,

Is the question complete? Could you clarify?

[emerald]

iFruit:

emerald:

Hi, pls help to solve:


(2h - 7k)(3k - 1)(3 - h)


Thanks.

Hi emerald,

Is the question complete? Could you clarify?

Hi iFruit,

This is the given question but maybe there's some problem with it cos my dd couldn't solve it. Anyway, thanks for responding.

[OK Lor]

Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.