Posted April 18, 2013 Find the smallest positive integer such that if the ones digit is moved (from the right) all the way to the left, the resulting number is exactly 50% more than the original number. 0 Share this post Link to post Share on other sites

0 Posted April 18, 2013 6digits? 5 7 1 4 2 8 -140000 -14000 -1400 -140 -14 99998.5 -700000 -98000 -1400 -560 -28 799988 = 0 0 Share this post Link to post Share on other sites

0 Posted April 18, 2013 The smalles positive integer is 2143 its 50% is 1071.5 result number is 3214.5 0 Share this post Link to post Share on other sites

0 Posted April 18, 2013 Let's the value be m and its shifted version n. 2/3*n = m m and n are both integers, so n-m=1/3*n is also an integer. Therefore 3 | n . As their digits have the same sum, 3 | m. Therefore 3 | n-m --> 3 | 1/3*n --> 9 | n. As their digits have the same sum, 9 | m. Following similar logic from above, above 9 | n-m --> 9 | 1/3*n --> 27 | n. So you need only account for 1/27 of the integers for a brute strength solution (which speeds it up significantly) I got 285714 and 571428. 0 Share this post Link to post Share on other sites

0 Posted April 18, 2013 Let's the value be m and its shifted version n. 2/3*n = m m and n are both integers, so n-m=1/3*n is also an integer. Therefore 3 | n . As their digits have the same sum, 3 | m. Therefore 3 | n-m --> 3 | 1/3*n --> 9 | n. As their digits have the same sum, 9 | m. Following similar logic from above, above 9 | n-m --> 9 | 1/3*n --> 27 | n. So you need only account for 1/27 of the integers for a brute strength solution (which speeds it up significantly) I got 285714 and 571428. brilliant observation 0 Share this post Link to post Share on other sites

Posted

## Share this post

## Link to post

## Share on other sites