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

 Doubtful Divisibility (Posted on 2012-05-09)
For two positive integers x and y, we have:
(100*x + y)/(x+y) = 1 + (99*x)/(x+y) and,
(100*x + y)/(x+y) = 100-(99*y)/(x+y) . . . . (i)

Since both x and y are positive, we have: x+y > x and x+y > y

Accordingly, x+y does not divide either x or y, and, consequently from (i) we deduce that x+y divides (100*x + y) whenever, (x+y) divides 99.

Now, substituting (x, y) = (20, 16), we have: x+y = 36 which does not divide 99 and consequently we can conclude that 36 does not divide 2016.

However, in reality 2016/36 = 56 and, accordingly the above result is flawed.

Can you spot the error?

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 2 of 3 |

"x+y divides (100*x + y) whenever, (x+y) divides 99" can be rephrased as "if (x+y) divides 99, then x+y divides (100*x + y)".  This is in fact, true.

However, the converse is not true. The error is in assuming that the converse of a true conditional must also be true.

 Posted by Charlie on 2012-05-09 13:40:51

 Search: Search body:
Forums (0)
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}