Heaven seven

Observe that $20020^{12} =2^{12} * 5^{12} * 7^{12} * 11^{12} * 13^{12}$ .

Suppose we have a factor with the desired property,If the factor is divisible by $2$ or $5$ then its residue modulo $10$ cannot be 7, hence the factor has a prime factorization of the form $7^l*11^m*13^n$ (where $0 \le l,m,n \le 12$

We now observe that $11^{n} = 1$ mod $10$ , and $7^l*13^n = 7$ mod $10$ if and only if one of the following four cases is true:

Case 1: $7^l = 1, 13^n =7$

$7$ and $13$ are relatively prime to $10$ , and $\phi (10) = 4$ , hence there are $4$ valid values of $l$ and $3$ of $n$ , and 12 $l,n$ combinations in this case.

Case 2: $7^l = 7, 13^n =1$

Similarly there are 12 $l,n$ combinations in this case.

Case 3: $7^l = 3, 13^n =9$

By the same reasoning as in case 1, there are $3$ valid values of $l$ and $3$ of $n$ , and 9 $l,n$ combinations overall.

Case 4: $7^l = 9, 13^n =3$

Similarly there are 9 $l,n$ combinations in this case.

Thus there are 42 $l,n$ combinations overall. Remembering that $m$ such that $0 \le m \le 12$ work for every $l,n$ combination, there are $42*13 = \boxed {546}$ such factors.

$20020 = 2^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ . The only factors that matter are $7, 11, 13$ .

We have 13 possibilities for 11 because it stays as 1.

Let $a$ be the exponent on 7 and $b$ be the exponent on 13. We have $a = 0; b = 3, a = 1, b = 0; a = 2, b = 1; a = 3, b = 2$ (in mod 4), contributing 12, 12, 9, and 9 solutions respectively. $13 * 42 = 546.$

First of all, we realize that any such divisors cannot have $2$ or $5$ as factors. So factoring those numbers out of $20020$ , we get $1001=7*11*13$ . Now consider, any such number will retain the same units digit if it is multiplied by $11$ . So we should see how many combinations of $13$ or less of $7s$ and $13s$ there are with a $7$ units digit. The first 13 multiples of 7 modulo 10 are ${1,7,9,3,1,7,9,3,1,7,9,3,1}$ and those of 13 are ${1,3,9,7,1,3,9,7,1,3,9,7,1}$ . As you can see, in both of them there are $3$ of [{3,7 and 9}) and $4$ ones. So, each 1 can be paired with a 7, each 3 with a 9, and each 7 with 1 and 9 with 3. Counting those, we get $42$ . Now, this must be multiplied by 13 possibilities for the factor of 11, and we get 546.

$20020 = 2^2 \times 5 \times 7 \times 11 \times 13$ therefore $20020^{12} = 2^{24} \times 5^{12} \times 7^{12} \times 11^{12} \times 13^{12}$ therefore $x$ is a divisor of $20020^{12}$ iff $x = 2^a \times 5^b \times 7^c \times 11^d \times 13^e$ , where $0 \le a \le 24$ and $0 \le b, c, d, e \le 12$ .

If $x$ ends with a 7, $a = 0$ , because otherwise it'd end in an even digit. For almost the same reason, $b = 0$ .

$d$ can have any value, because multiplying a number by $11^n$ doesn't change the number's units digit.

Now we're looking for values of $c$ and $d$ . Since $x^y (mod 10)$ is periodic, following pairs of values are suitable:

• $c = 4k, d = 4l + 3$
• $c = 4k + 1, d = 4l$
• $c = 4k + 2, d = 4l + 1$
• $c = 4k + 3, d = 4l + 2$

where $k$ and $l$ are non-negative.

(make a table of the last digits of $7^x$ and $13^x$ if this is unclear).

Counting how many suitable pairs we have, we get $4 \times 3 + 3 \times 4 + 3 \times 3 + 3 \times 3 = 42$ . Remembering that we have 13 possible values for $d$ , the answer is $42 \times 13 = 546$ .

prime factorization of your number is N = 20020¹² = 2²⁴ × 5¹² × 7¹² × 11¹² × 13¹².

So, the set of all divisors of N would be D = {2ᵃ×5ᵇ×7ᶜ×11ᵈ×13ᵉ | a, b, c, d, e ϵ ℕ, a ≤ 24, b, c, d, e ≤ 12}.

And since each assignment of values to a, b, c, d, e gives a unique number, there would be a total of (24+1)×(12+1)×(12+1)×(12+1)×(12+1) = 714025 different divisors of N.

But to get a 7 in the units digit, no divisor should be a multiple of 2 or 5. Removing them, we have the set of divisors as D' = {7ᵃ×11ᵇ×13ᶜ | a, b, c ϵ ℕ, a, b, c ≤ 12}, which are (12+1)×(12+1)×(12+1) = 2179 in number. But not all such divisors will have 7 as their units digit. So we'd analyse a bit more.

We need not worry about the factor 11, since it does not affect the units digit. The problem boils down to finding all multiples formed by combinations of 7 and 13, such that the units digit is 7. If we only look only at the units digit, what combinations of 7 and 3 give us 7? Thus, we need to find S = {(a, b) | u(3ᵃ×7ᵇ) = 7, a, b ϵ ℕ, a, b ≤ 12}, where u(·) means "units digit".

Now, units digit, u(3ᵃ×7ᵇ) = 7 only in the following cases: Case 1: u(3ᵃ) = 1 and u(7ᵇ) = 7 Case 2: u(3ᵃ) = 7 and u(7ᵇ) = 1 Case 3: u(3ᵃ) = 3 and u(7ᵇ) = 9 Case 4: u(3ᵃ) = 9 and u(7ᵇ) = 3

But, since a, b ≤ 12, we have: u(3ᵃ) = 1 if and only if a ϵ {0, 4, 8, 12} u(7ᵇ) = 1 if and only if b ϵ {0, 4, 8, 12} u(3ᵃ) = 7 if and only if a ϵ {3, 7, 11} u(7ᵇ) = 7 if and only if b ϵ {1, 5, 9} u(3ᵃ) = 3 if and only if a ϵ {1, 5, 9} u(7ᵇ) = 3 if and only if b ϵ {3, 7, 11} u(3ᵃ) = 9 if and only if a ϵ {2, 6, 10} u(7ᵇ) = 9 if and only if b ϵ {2, 6, 10}

Thus, For case 1, we have a ϵ {0, 4, 8, 12} and b ϵ {1, 5, 9}; 4 × 3 = 12 For case 2, we have a ϵ {3, 7, 11} and b ϵ {0, 4, 8, 12}; 3 × 4 = 12 For case 3, we have a ϵ {1, 5, 9} and b ϵ {2, 6, 10}; 3 × 3 = 9 For case 4, we have a ϵ {2, 6, 10} and b ϵ {3, 7, 11}; 3 × 3 = 9 having 12, 12, 9, and 9 feasible combinations respectively. So, we have a total of 12+12+9+9 = 42 combinations. Now, considering the 13 possible multiples of 11 (including 11⁰, 11¹, ..., 11¹²), we get 42 × 13 = 546 divisors altogether, which have last digit as 7. Thus, the answer is 546.

First, it is found that $20020^{12}$ = $2^{24}$ $\times$ $5^{12}$ $\times$ $7^{12}$ $\times$ $11^{12}$ $\times$ $13^{12}$ . Note that the divisors 2, 5 and 11 do not yield 7 as the unit digit if they are used as factors of a number N. Let the positive divisors that have 7 as their unit digits be N. There are 6 cases that fulfill the given condition:

Case 1: When N is in the form $7^{a}$ , there are 3 powers of 7 that has 7 as their unit digits, namely $7^{1}$ , $7^{5}$ and $7^{9}$ .

Case 2: Similarly, when N is in the form $13^{b}$ , there are 3 possible values of N.

Case 3: When N is in the form $7^{c}$ $\times$ $11^{d}$ , the unit digit of powers of 7 are 7, 9, 3, 1 which "rotates" while powers of 11 only have 1 as its unit digit, regardless of its power. Take note that only 7 $\times$ 1 = 7. The powers of 7 that yields 7 as its unit digit are $7^{1}$ , $7^{5}$ and $7^{9}$ . There are 3 possible values of c and 12 possible values of d that yield 7 as the unit digit for $7^{c}$ $\times$ $11^{d}$ . Hence, there are 3 $\times$ 12 = 36 values of N.

Case 4: When N is in the form $7^{e}$ $\times$ $13^{f}$ , the unit digits of powers of 7 are as stated above, and the unit digits of powers of 13 are 3, 9, 7 and 1. In this case, 9 $\times$ 3, 3 $\times$ 9, 1 $\times$ 7 and 7 $\times$ 1 yields 7 as the unit digit. Consider several conditions:

1. 9 $\times$ 3: To yield 9 as the unit digit of $7^{e}$ , there are 3 values of e. To yield 3 as the unit digit of $13^{f}$ , there are 3 values of f. Hence, there are 3 $\times$ 3 = 9 values of N fulfilling Condition 1.

2. 3 $\times$ 9: Similar to Condition 1, to yield 3 as the unit digit of $7^{e}$ , there are 3 values of e. To yield 9 as the unit digit of $13^{f}$ , there are 3 values of f. Hence, there are 3 $\times$ 3 = 9 values of N fulfilling Condition 2.

Same goes to Condition 3 and 4, namely 1 $\times$ 7 and 7 $\times$ 1 respectively. These two conditions share the same number of values of N too, i.e. 9. Thus, there are a total of 9 $\times$ 4 = 36 values of N in Case 4.

Case 5: Similar to Case 3, when N is in the form $11^{g}$ $\times$ $13^{h}$ , the unit digit of powers of 13 are 3, 9, 7 and 1 which "rotates" while powers of 11 only have 1 as its unit digit, regardless of its power. Hence, only 7 $\times$ 1 = 7. The powers of 13 that yields 7 as its unit digit are $13^{3}$ , $13^{7}$ and $13^{11}$ . There are 12 possible values of g and 3 possible values of h that yield 7 as the unit digit for $11^{g}$ $\times$ $13^{h}$ . Hence, there are 3 $\times$ 12 = 36 values of N.

Case 6: N is in the form $7^{j}$ $\times$ $11^{k}$ $\times$ $13^{m}$ . We have known the unit digits of powers of 7, 11 and 13. Among these numbers, there are 4 possible combinations of unit digits of powers of 7, 11 and 13 of which their products have 7 as their unit digits, namely (7,1,1), (9,1,3), (3,1,9) and (1,1,7) respectively. Consider several conditions:

1. (7,1,1): The value of j should be 1, 5, 9 and the value of m should be 4, 8, 12. The value of k ranges from 1 to 12. Thus, there are 3 $\times$ 12 $\times$ 3 = 108 values of N fulfilling this condition.

Condition 2, 3 and 4, namely (9,1,3), (3,1,9) and (1,1,7) also yields the same number of N, i.e. 108 for each condition. Thus, for Case 6, there are altogether 108 $\times$ 4 = 432 values of N.

Total number of N = Case 1 + Case 2 + Case 3 + Case 4 + Case 5 + Case 6 = 3 + 3 + 36 + 36 + 36 + 432 = 546, which is what we want to get.

We factorize: $20020=2^2\cdot 5\cdot 7\cdot 11\cdot 13$ An efficient way to do this is to recognize the factor of 1001, which I happen to remember (probably because it is 1001) is $7\cdot 11\cdot 13$ .

If the number ends in 7, we cannot take 2 or 5. Then, we are left with $7^{12}\cdot 11^{12}\cdot 13^{12}$ We can take any number of 11's without affecting the units digit, so we multiply by 13 at the end.

We can write down the last digits of powers of 7 and 13: $7,9,3,1,7,9,3,1,7,9,3,1$ $3,9,7,1,3,9,7,1,3,9,7,1$ Now, we divide into cases:

Case 1: We take at least one 7 and at least one 13

We notice that for each power of 7, there are 3 powers of 13 that match: $7\cdot 1=7$ $9\cdot 3=27$ $3\cdot 9=37$ $1\cdot 7=7$ Therefore, there are $12\cdot 3=36$ possibilities here.

Case 2: Either a power of 7 or a power of 13

In this case, we must take a power that ends in 7. There are $6$ .

The final answer is $13\cdot(36+6)=\boxed{546}$ .