k-phobic Numbers
A five-digit positive integer is called -phobic if no matter how one chooses to alter at most four of the digits, the resulting number (after disregarding any leading zeroes) will not be a multiple of . Find the smallest positive integer value of such that there exists a -phobic number.
The answer is 11112.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
When k ≤ 1 0 0 0 0 , each of the intervals [ 1 0 0 0 0 , 1 9 9 9 9 ] , [ 2 0 0 0 0 , 2 9 9 9 9 ] , . . . [ 9 0 0 0 0 , 9 9 9 9 9 ] contains a multiple of 9 . Keep in mind that each interval contains 1 0 0 0 0 consecutive integers.
If we consider 1 0 0 0 0 < k ≤ 1 1 1 1 1 , there exists a multiple of k with any leading digit, because these intervals contain k , 2 k , . . . , 9 k . Therefore, there exists a k -phobic number, since we can keep the leading digit and change everything to a multiple of k
When k = 1 1 1 1 2 , the only multiples we can think of in the range are:
0 0 0 0 0 , 1 1 1 1 2 , 2 2 2 2 4 , 4 4 4 4 8 , 3 3 3 3 6 , 5 5 5 6 0 , 7 7 7 8 4 , 8 8 8 9 6 , 9 9 9 5 9 .
Thus, 1 1 1 1 2 is the smallest number that is required.