12
\$\begingroup\$

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.
Improve this question
\$\endgroup\$
4
  • \$\begingroup\$ Does 667 (=23*29) make for two 2s, one 3, and one 9 in step 3? \$\endgroup\$ Commented Dec 23, 2017 at 18:10
  • \$\begingroup\$ @JonathanAllan Yes. \$\endgroup\$ Commented Dec 23, 2017 at 18:19
  • 2
    \$\begingroup\$ @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. \$\endgroup\$ Commented Dec 23, 2017 at 18:26
  • \$\begingroup\$ @JonathanAllan It would be nice if I could count. :-/ \$\endgroup\$ Commented Dec 23, 2017 at 18:36

6 Answers 6

Reset to default
4
\$\begingroup\$

Jelly,  18  17 bytes

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

DF‘ċЀ⁵ ÆfQÇæ.Ç‘$

A monadic link taking a positive integer and returning a non-negative integer.

Try it online!

How?

DF‘ċЀ⁵ - Link 1, digitalCount: number(s) e.g. [13,17] D - to decimal list (vectorises) [[1,3],[1,7]] F - flatten [1,3,1,7] ‘ - increment (vectorises) [2,4,2,8] ⁵ - literal ten 10 Ѐ - map across (implicit range [1,2,3,4,5,6,7,8,9,10]) ċ - count [0,2,0,1,0,0,0,1,0,0] ÆfQÇæ.Ç‘$ - Main link: positive integer, n e.g. 11999 $ - 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
Improve this answer
\$\endgroup\$
2
  • \$\begingroup\$ 17 bytes \$\endgroup\$ Commented Dec 23, 2017 at 19:34
  • \$\begingroup\$ Or use dot product \$\endgroup\$ Commented Dec 23, 2017 at 19:45
3
\$\begingroup\$

Jelly, 16 bytes

ṾċЀØD 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.

ṾċЀØD 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'] ċЀ 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
Improve this answer
\$\endgroup\$
2
\$\begingroup\$

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

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

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)
Improve this answer
\$\endgroup\$
1
\$\begingroup\$

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!

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Python 2, 136 127 bytes

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

Try it online!

Credits

Improve this answer
\$\endgroup\$
3
  • \$\begingroup\$ 127 bytes \$\endgroup\$ Commented Dec 23, 2017 at 20:42
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Dec 23, 2017 at 20:45
  • \$\begingroup\$ You can take a look through this for -~ and related stuff. \$\endgroup\$ Commented Dec 23, 2017 at 20:48

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.