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Pinaki Biswas 6 months, 1 week ago.
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Day 1 problem
Time limit-3 days
Problem no 1.
Find the smallest positive integer n such that n! ends in 290 “0”s.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.
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. 🙂
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