pythagoras123
  • 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.
Mathematics
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SOLVED
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chestercat
  • chestercat
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experimentX
  • experimentX
what are zeros??
pythagoras123
  • 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
  • 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
  • experimentX
@ffm i'll try before you
pythagoras123
  • pythagoras123
Just tell me how you work it out, or the solution/methodology. The answer is not so important.
experimentX
  • experimentX
i think between 1 and 10, you will have 2 zeros.
anonymous
  • 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
  • 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
  • experimentX
sorry, 139
experimentX
  • experimentX
@ffm is it correct??
anonymous
  • anonymous
33 for 139.
anonymous
  • anonymous
I have posted the correct answer.
experimentX
  • experimentX
ah ... where did i screwed up??
pythagoras123
  • pythagoras123
@FoolforMath, how did you manage to get 124?
experimentX
  • experimentX
looks like my theory screwed up from here http://www.wolframalpha.com/input/?i=30%21%2F20%21&dataset=&equal=Submit
experimentX
  • 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
  • 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
  • experimentX
the answer was n = 120
anonymous
  • anonymous
The largest power of 5 dividing n must be 5^28.Now use De Polignac's Formula.
anonymous
  • anonymous
120! = 6689502913449127057588118054090372586752746333138029810295671352301633557244962989366874165271984981308157637893214090552534408589408121859898481114389650005964960521256960000000000000000000000000000 Of course you need an analytic proof of that using 2 and 5.
anonymous
  • anonymous
Should the question be, find the smallest n to achieve that?

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