Find remainder: [22222...(1000times)]/[5^5]. Please explain the method also. Thanks?

Question

its divided by 5^5, i think its partially cutoff frm my question.

Answer 1

we know 10^5 is divisible by 5^5

now 2222..... (1000) times mod 10^5 = 22222

so 2222..... (1000) times mod 5^5 = 22222 mod 5^5

= 22222 mod 5^5 = 22222 mod 3125 = 347

thus 347 is the remainder

Answer 2

The remainder is 347.

Reasoning:
The large number is equal to:
2*10^0 + 2*10^1 + 2*10^2 + ... + 2*10^999
Another way to look at that is to group the first 5 terms together like this:
22222 + 2*10^5 + 2*10^6 + ... + 2*10^999

All of the terms from 2*10^6 onward have a curious property. They do not have any effect on the remainder of the division!
Here's why. Let's look at the prime factors of one of those terms. The nth term is:
2*10^n
Its factors are: 2 (n+1 times) and 5 (n times)
that is: 2*10^n = 2^(n+1) * 5^n
Whenever n >= 5, this number must be evenly divisible by 5^5. So, those terms do not affect the remainder.

All that is left to do now is to find the remainder of 22222 / 5^5.
That remainder is 347.

Answer 3

Let's write 222...(1000) times as
2*10^999 + 2*10^998 + ... + 2*10^6 + 22222.
Also, 5^5 = 3125.
Since 10^6 is divisible by 10^5 it is also divisible
by 5^5, so all terms in this expression except
the last are divisible by 3125.
So the answer is congruent to 22222 mod(3125).
Next, 22222= 2(10^4 + 10^3 + 10^2 + 10 + 1)
and 10^4 = 625(mod 3125)
So our final remainder is
2(625 + 1000 +111) = 2(1625 + 111) = 2(1736) = 3472 = 347(mod 3125)

Answer 4

Thus 5^5 = 3125; 22222 =7*3125 +347; 222222 =71*3125 +347; 2222222 =711*3125 +347; and so on - induction; remainder = 347; because N= 200000 /3125 = 64 is integer, hence N*10^k /3125 = 64*10^k is also integer;