Divisors with units digit 5

$20130 = 2 \times 3 \times 5 \times 11 \times 61$ , so $20130 ^ {4}$ has the same prime divisors, everyone of them 4 times. If we want to count the divisors with the unit digit 5, we need to count the divisors which are divisible by 5 and not by 2. So their prime divisors must be 5, and all combinations (and the empty one, too) with the three other 5's, four 3's, four 11's and four 61's. There are four options for the power of 5- 1,2,3 or 4. For the other prime divisors there are five options for the power- 0,1,2,3 or 4.

So there are $4 \times 5 \times 5 \times 5 = \boxed {500}$ divisors with the unit digit 5.

First, we prime factorize the expression. $20130^4=2^4\cdot3^4\cdot5^4\cdot11^4\cdot61^4$ If the divisor must end in 5, we cannot have any factors of 2 and at least one factor of 5. Thus, we have 1 option for the number of factors of 2 (0 factors), 5 options for the number of factors of 3 (0, 1, 2, 3, or 4 factors), 4 options for the number of factors of 5 (1, 2, 3, or 4 factors), 5 options for the number of factors of 11 (0, 1, 2, 3, or 4 factors), and 5 options for the number of factors of 61 (0, 1, 2, 3, or 4 factors). Therefore, our expression for the number of positive divisors that fit the requirement is $1\cdot5\cdot4\cdot5\cdot5=\boxed{500}$

$20130^4=(2\cdot3\cdot5\cdot11\cdot 61$

$2$ shouldn't be a factor of the divisor, so the divisors are a combination of $3^4, 5^4, 11^4, 61^4$

Total number of such factors are : $(4+1)(4+1)(4+1)(4+1)(4+1)=625$ .

Here, we have also counted the cases where $5$ is not a factor. The number of such cases is the combination of the factors $3^4, 11^4, 61^4$ , which is $125$ . Hence the answer : $625-125=\boxed{500}$ .

20130 =2013 x 10 = (61 x 3 x 11) x (2 x 5)

so (20130)^4 = 61^4 . 3^4 . 11^4 . 5^4 . 2^4

thus we can see the divisors of (20130)^4 are of the form:

==>Divisor =(61^a).(3^b).(11^c).(5^d).(2^e) , --------->where 0≤ {a,b,c,d,e,f} ≤ 4

we know 5 x (any odd number) results a 5 in units digit place

so our specific divisors are odd multiples of 5 ====> e=0, 1≤d≤4

so 0≤{a,b,c}≤4 --------> a,b,c can assume a value in 5 ways ,ie {0,1,2,3,4}

1≤d≤4-------->d can assume any value in 4 ways, ie{1,2,3,4}

e=0 ----->e has only one choice

so total ways of describing our specific divisors,N = (ways possible for 'a') x(ways posssible for 'b')x....x(ways possible for 'e')

===>N = 5 x 5 x 5 x 4 x 1 = 500 is the soln.

$20130^4=61^4\times11^4\times5^4\times3^4\times2^4$ . As the numbers ending with $5$ are some odd number multiplied by $5$ , the number of divisors of it ending with $5$ is equal to the number of divisors of $61^4\times11^4\times5^3\times3^4$ .

$20130 = 2 * 3 * 5 * 11 * 61$ . If we want a divisor ending with $5$ , then the factor of $5$ should be multiplied by an odd number. So we work with the factors $3*5*11*61$ . As the problem says $20130^{4}$ , then we'll work with $3^{4}*5^{4}*11^{4}*61^{4}$ . But remember that we need to include the $1$ , so $3^{0}$ , $11^{0}$ and $61^{0}$ count. So the question can be transformed: How many multiplications on $3^{4}*5^{4}*11^{4}*61^{4}$ end with $5$ ? We can take $5$ powers of $3$ , $4$ powers of $5$ , $5$ powers of $11$ and $5$ powers of $61$ . The answer is $5*4*5*5 = \boxed {500}$ .