Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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.
 2 years ago
 2 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.
 2 years ago
 2 years ago

This Question is Closed

pythagoras123Best 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)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
It's somewhat tedious to explain the whole thing. One interesting observation is there would be only trailing zeroes.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
@ffm i'll try before you
 2 years ago

pythagoras123Best ResponseYou've already chosen the best response.0
Just tell me how you work it out, or the solution/methodology. The answer is not so important.
 2 years ago

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

FoolForMathBest ResponseYou've already chosen the best response.1
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
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
"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.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
@ffm is it correct??
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
I have posted the correct answer.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
ah ... where did i screwed up??
 2 years ago

pythagoras123Best ResponseYou've already chosen the best response.0
@FoolforMath, how did you manage to get 124?
 2 years ago

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

experimentXBest ResponseYou've already chosen the best response.0
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.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
110 : 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
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
the answer was n = 120
 2 years ago

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

eliassaabBest ResponseYou've already chosen the best response.1
120! = 6689502913449127057588118054090372586752746333138029810295671352301633557244962989366874165271984981308157637893214090552534408589408121859898481114389650005964960521256960000000000000000000000000000 Of course you need an analytic proof of that using 2 and 5.
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.1
Should the question be, find the smallest n to achieve that?
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.