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TASK #1 - Number Divisors
Write a script to find lowest 10 positive integers having exactly 8 divisors.
Example
24 is the first such number having exactly 8 divisors.
1, 2, 3, 4, 6, 8, 12 and 24.
MY NOTES: Very easy.
TASK #2 - Like Numbers
You are given positive integers, $m and $n. Write a script to find
total count of integers created using the digits of $m which is also
divisible by $n.
Repeating of digits are not allowed. Order/Sequence of digits can't
be altered. You are only allowed to use (n-1) digits at the most. For
example, 432 is not acceptable integer created using the digits of
1234 (because the digits are in the wrong order). Also for 1234, you can
only have integers having no more than three digits.
Example 1:
Input: $m = 1234, $n = 2
Output: 9
Possible integers created using the digits of 1234 are:
1, 2, 3, 4, 12, 13, 14, 23, 24, 34, 123, 124, 134 and 234.
There are 9 integers divisible by 2 such as:
2, 4, 12, 14, 24, 34, 124, 134 and 234.
Example 2:
Input: $m = 768, $n = 4
Output: 3
Possible integers created using the digits of 768 are:
7, 6, 8, 76, 78 and 68.
There are 3 integers divisible by 4 such as:
8, 76 and 68.
MY NOTES: Sounds pretty easy. However, there's one mistake:
"You are only allowed to use (n-1) digits at the most"
is flatly contradicted by the examples. I assume it meant
"length(m)-1" which is what the examples show.
So each digit can either be present or absent - it's another "subsets"
task. One wrinkle: the length(m)-1 rule means that (all-present) is
not a valid combination. Plus of course (all-absent) isn't either.
Last time we did a "subsets" task was in Challenge 136 Task 2,
Fibonacci Seqence. In that, I wrote:
"How do we sum subsets of values? Two obvious ways:
1. recursive function, iterating over the values: each value is either in the
subset or not, try both possibilities.
2. binary-counting method, iterate over every combination C from 1 ..
2**(number values)-1, then sum up only the elements where the corresponding
bit in C is on.
"
That time, I wrote a (2) binary-counting method. So this time round, let's
try a recursive function instead, just for variety.
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