# Primes ’n’ Digits

This has no practical purpose but it could be fun to golf.

# Challenge

Given a number n,

1. Count the amount of each digit in n and add 1 to each count
2. Take the prime factorization of n
3. Count the amount of each digit in the prime factorization of n, without including duplicate primes
4. Create a new list by multiplying together the respective elements of the lists from steps 1 and 3
5. Return the sum of that list

For example, 121 has two 1s and a 2, so you would get the following list from step 1:

0 1 2 3 4 5 6 7 8 9
1 3 2 1 1 1 1 1 1 1


The prime factorization of 121 is 112, which gives the following list for step 3:

0 1 2 3 4 5 6 7 8 9
0 2 0 0 0 0 0 0 0 0


Note how we did not count the exponent. These multiply together to get:

0 1 2 3 4 5 6 7 8 9
0 6 0 0 0 0 0 0 0 0


And the sum of this list is 6.

# Test cases

1 -> 0
2 -> 2
3 -> 2
4 -> 1
5 -> 2
10 -> 2
13 -> 4
121 -> 6


# Notes

• Standard loopholes are forbidden.
• Input and output can be in any reasonable format.
• You should leave ones (or zeros for step 3) in the list for digits that did not appear in the number.
• This is , so the shortest solution in bytes wins.
• Does 667 (=23*29) make for two 2s, one 3, and one 9 in step 3? Dec 23 '17 at 18:10
• @JonathanAllan Yes. Dec 23 '17 at 18:19
• @wizzwizz4 232792560 -> [2,1,4,2,1,2,2,2,1,2] (step 1); 2*2*2*2*3*3*5*7*14*17*19 (step 2); so [0,5,1,2,0,1,0,2,0,1] (step 3); then [0,5,4,4,0,2,0,4,0,2] (Step 4); and hence should output 21. Dec 23 '17 at 18:26
• @JonathanAllan It would be nice if I could count. :-/ Dec 23 '17 at 18:36

# Jelly,  18  17 bytes

-1 byte thanks to caird coinheringaahing & H.PWiz (avoid pairing the two vectors)

DF‘ċÐ€⁵
$- last two links as a monad: Ç - call the last link (1) as a monad [0,2,0,0,0,0,0,0,0,3] ‘ - increment (vectorises) [1,3,1,1,1,1,1,1,1,4] Æf - prime factorisation [13,13,71] Q - deduplicate [13,17] Ç - call the last link (1) as a monad [0,2,0,1,0,0,0,1,0,0] æ. - dot product 8  • 17 bytes Dec 23 '17 at 19:34 • Or use dot product Dec 23 '17 at 19:45 # Jelly, 16 bytes ṾċÐ€ØD ÆfQÇ×Ç‘$S


Try it online!

Developed independently from and not exactly the same as the other Jelly solution.

Explanation

I'm gong to use 242 as an example input.

ṾċÐ€ØD     Helper link
Ṿ          Uneval. In this case, turns it's argument into a string.
242Ṿ → ['2','4','2']. [2,11] → ['2', ',', '1', '1']. The ',' won't end up doing anything.
ØD     Digits: ['0','1',...,'9']
ċÐ€       Count the occurrence of €ach digit in the result of Ṿ

ÆfQÇ×Ç‘$S Main link. Argument 242 Æf Prime factors that multiply to 242 → [2,11,11] Q Unique elements → [2,11] Ç Apply helper link to this list → [0,2,1,0,0,0,0,0,0,0] Ç‘$   Apply helper link to 242 then add 1 to each element → [1,1,3,1,2,1,1,1,1,1]
×      Multiply the two lists element-wise → [0,2,3,0,0,0,0,0,0,0]
S  Sum of the product → 5


# APL (Dyalog), 43 41 bytes

⎕CY'dfns'
+/×/+/¨⎕D∘.=⍕¨(⎕D,r)(∪3pco r←⎕)


Try it online!

How?

r←⎕ - input into r

3pco - prime factors

∪ - unique

⎕D,r - r prepended with 0-9

⍕¨ - format the factors and the prepended range

⎕D∘.= - cartesian comparison with every element of the string 0123456789

+/¨ - sum each row of the two tables formed

×/ - multiply the two vectors left

+/ - sum the last vector formed

# Pip, 44 bytes

Y_N_.aM,tT++o>aTa%o{a/:olPBo}\$+y*Y_N JUQlM,t


Takes input from command-line argument. Try it online!

# Python 2, 136 127 bytes

lambda a:sum(''.join(u(a)).count(i)*-~a.count(i)for i in range(10))
u=lambda a:[jfor j in range(2,a)if a%j<1>len(u(j))]


Try it online!

# Credits

• 127 bytes Dec 23 '17 at 20:42
• @Mr.Xcoder Updated, thanks for showing me the use of -~ I was always a bit confused on that. And I need to start remembering the <1 thing. Thanks for the help.
– Neil
Dec 23 '17 at 20:45
• You can take a look through this for -~ and related stuff. Dec 23 '17 at 20:48

# 05AB1E (legacy), 5 bytes

fS¢>O


Try it online!

f      list of prime factors (without duplicates) of the implicit input
S      characters, all of the digits
¢      count each of the characters in the implicit input
>      increase each of the counts
O      sum (implicit output)