Posted 18 Apr 2013 · Report post 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 18 Apr 2013 · Report post 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 18 Apr 2013 · Report post 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 18 Apr 2013 · Report post 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 18 Apr 2013 · Report post 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

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