Math Olympiad, I.S.I. Entrance and College Mathematics › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Combinatorics › Problem Solving Marathon for \"Thousand flowers\"

- AuthorPosts
- January 7, 2019 at 8:39 pm #25311
Day 1 problem

Time limit-3 days

Problem no 1.

Find the smallest positive integer n such that n! ends in 290 “0”s.January 8, 2019 at 11:39 am #25397Answer is 1170.

Zero in end digits comes when we multiply 5 with even numbers …

1! = 1 = 1

2! = 1*2 = 1

3! = 1*2*3 = 6

4! = 1*2*3*4 = 24

5! = 1*2*3*4*5 = 120

120 has one zero and n has one five..

same lines if we move forward next additional zero will come when we reach 10!

10! will have two end digit zero.

So we need to find n which gives us 290 end zero.

check with 100 ! – we will get 24 5’s & hence 24 zeros …which is less than expected 290 zeros..

move with 1000!

1000 has 249 5’s

so we move forward say 1200 has 298 5’s and hence 298 zero.

1150 has 286 5’s and hence 286 zero.

On same lines if we move we will get 1170.

January 8, 2019 at 8:04 pm #25409There are 2 zeros in 10! as, 2*5 has one zero and 10*1 has another.

Using the same logic, we will get that 20! has 4 zeros.

However, 30! has 7 zeros because 5*5=25.

So, we have to look out for 5^x and multiples of them.

40! has 9 zeros.

50! has 11 + 1 zeros because 50=2*5^2.

60! has 14 zeros.

70! has 16 zeros.

80! has 18+1 zeros as 75=3*5^2.

90! has 21 zeros.

100! has 23+1 zeros =24 as 100=4*5^2.

We can find an algorithm:100/5=20

100/5^2=4

20+4=24; It is the number of zeros in 100!.

So now we have a powerful weapon to solve this problem.

No. of zeros in 1000!:1000/5=200

1000/5^2=40

1000/5^3=8

1000/5^4=1.6(We will consider the whole no. part only)

So no. of zeros in 1000!=249.

Similarly no. of zeros in 1100!=220+44+8+1=273.

1200!=240+48+9+1=298

1150!=230+46+9+1=286

1175!=235+47+9+1=292

1170!=234+46+9+1=290.

We got our answer. It is

**1170.**🙂 - AuthorPosts

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