All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Insert Zero (Posted on 2016-03-21)

Find all integers n greater than nine that satisfy
the following:

The integer formed by inserting the zero digit
between the units and the tens digits of n is
a multiple of n.

 See The Solution Submitted by Bractals No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 2 of 5 |
Let the original n be 10x+y and the new number m be 100x+y. x must be a positive integer and y must be an integer 0-9.

Then (100x+y)/(10x+y) = 10 - (9y)/(10x+y) is an integer.

If y=0 then the fraction is 0 for all x.  So the set of all n includes all multiples of 10.

If y=1,2,3,4,6,7,9 then there are no such positive integer x to make (9y)/(10x+y) an integer.

If y=5 then x=1 yields 45/15 = 3.  n=15, m=105 is a solution with 105/15 = 10-3.

If y=8 then x=1 yields 72/18 = 4.  n=18, m=108 is a solution with 108/18 = 10-4.

So the set of all n consists of 15, 18, and all multiples of 10.

 Posted by Brian Smith on 2016-03-21 12:25:37

 Search: Search body:
Forums (0)