## pythagoras123 3 years ago 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.

1. experimentX

what are zeros??

2. 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)

3. FoolForMath

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

4. experimentX

@ffm i'll try before you

5. pythagoras123

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

6. experimentX

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

7. FoolForMath

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

8. FoolForMath

"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.

9. experimentX

sorry, 139

10. experimentX

@ffm is it correct??

11. FoolForMath

33 for 139.

12. FoolForMath

I have posted the correct answer.

13. experimentX

ah ... where did i screwed up??

14. pythagoras123

@FoolforMath, how did you manage to get 124?

15. experimentX

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

16. 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.

17. 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

18. experimentX

the answer was n = 120

19. Aron_West

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

20. eliassaab

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

21. eliassaab

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