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anonymous
 4 years ago
The product of n whole numbers
1 x 2 x 3 x 4 x ... x (n1) x n, has twentyeight consecutive zeros. Find the largest value of n.
anonymous
 4 years ago
The product of n whole numbers 1 x 2 x 3 x 4 x ... x (n1) x n, has twentyeight consecutive zeros. Find the largest value of n.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"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
 4 years ago
Best ResponseYou've already chosen the best response.0It's somewhat tedious to explain the whole thing. One interesting observation is there would be only trailing zeroes.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0@ffm i'll try before you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just tell me how you work it out, or the solution/methodology. The answer is not so important.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i think between 1 and 10, you will have 2 zeros.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0"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
 4 years ago
Best ResponseYou've already chosen the best response.0@ffm is it correct??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have posted the correct answer.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0ah ... where did i screwed up??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolforMath, how did you manage to get 124?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0looks like my theory screwed up from here http://www.wolframalpha.com/input/?i=30%21%2F20%21&dataset=&equal=Submit

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.03 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
 4 years ago
Best ResponseYou've already chosen the best response.0110 : 2 zeros 1020: 2 2030: 3 (25 is factor of 100) 3040: 2 4050: 3 (50 is factor of 100) 6060: 2 6070: 2 7080: 3 (at some point 75 will give two zeros) 8090: 2 90100: 3 (100 itself gives two)  so from 0 to 100 we have: 24 zeros 100110: 2 110120: 2  I guess that would give 28 zeros screw you > minions of 100

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0the answer was n = 120

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The largest power of 5 dividing n must be 5^28.Now use De Polignac's Formula.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0120! = 6689502913449127057588118054090372586752746333138029810295671352301633557244962989366874165271984981308157637893214090552534408589408121859898481114389650005964960521256960000000000000000000000000000 Of course you need an analytic proof of that using 2 and 5.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Should the question be, find the smallest n to achieve that?
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