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- pythagoras123

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

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- pythagoras123

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- experimentX

what are zeros??

- pythagoras123

"0". This is a zero.
Consecutive zeros are zeros that appear consecutively, i.e. The product of 10 x 10x 10 has 3 consecutive zeros (1ooo)

- anonymous

It's somewhat tedious to explain the whole thing. One interesting observation is there would be only trailing zeroes.

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- experimentX

@ffm i'll try before you

- pythagoras123

Just tell me how you work it out, or the solution/methodology. The answer is not so important.

- experimentX

i think between 1 and 10, you will have 2 zeros.

- anonymous

I have used binary search along with the usual De Polignac's formula (just) twice to yield 124 as the answer.
Ref: http://en.wikipedia.org/wiki/De_Polignac%27s_formula

- anonymous

"Consecutive zeros are zeros that appear consecutively, i.e. The product of 10 x 10x 10 has 3 consecutive zeros (1ooo)"
You haven't understood the problem.

- experimentX

sorry, 139

- experimentX

@ffm is it correct??

- anonymous

33 for 139.

- anonymous

I have posted the correct answer.

- experimentX

ah ... where did i screwed up??

- pythagoras123

@FoolforMath, how did you manage to get 124?

- experimentX

looks like my theory screwed up from here
http://www.wolframalpha.com/input/?i=30%21%2F20%21&dataset=&equal=Submit

- experimentX

3 zeros ,,,?? how
25*22 = 00 => two zeros were coming from here.
so all the factors of 100 or 1000 or 10000 ... were causing problem for my idea.

- experimentX

1-10 : 2 zeros
10-20: 2
20-30: 3 (25 is factor of 100)
30-40: 2
40-50: 3 (50 is factor of 100)
60-60: 2
60-70: 2
70-80: 3 (at some point 75 will give two zeros)
80-90: 2
90-100: 3 (100 itself gives two)
-----------------------
so from 0 to 100 we have: 24 zeros
100-110: 2
110-120: 2
-----------------------
I guess that would give 28 zeros
screw you ---> minions of 100

- experimentX

the answer was n = 120

- anonymous

The largest power of 5 dividing n must be 5^28.Now use De Polignac's Formula.

- anonymous

120! = 6689502913449127057588118054090372586752746333138029810295671352301633557244962989366874165271984981308157637893214090552534408589408121859898481114389650005964960521256960000000000000000000000000000
Of course you need an analytic proof of that using 2 and 5.

- anonymous

Should the question be, find the smallest n to achieve that?

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