last updated by  Pinaki Biswas 2 weeks ago
3 voices
2 replies
  • Author
    Posts
  • #25311
    Udita Chatterjee
    Moderator

    Day 1 problem

    Time limit-3 days

    Problem no 1.
    Find the smallest positive integer n such that n! ends in 290 “0”s.

    #25397
    Saikrish Kailash
    Participant

    Answer 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.

     

     

     

     

    #25409
    Pinaki Biswas
    Participant

    There 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. 🙂

     

     

     

     

     

     

     

     

     

     

     

You must be logged in to reply to this topic.