bonanova Posted November 16, 2016 Report Share Posted November 16, 2016 Inspired by FUZZY's recent puzzle Suppose abcd is the normal decimal representation of a number: abcd = (1000a)+(100b)+(10c)+c. The digital sum of a number is defined as a+b+c+d. Are there numbers for which the product of its digital sum and its reversal is the original number? For example, the digital sum of 12345 is 15. Its reversal is 51. 15x51 is 765, 12345 is not equal ti 765, So 12345 is not a solution. Hint: 81 is a solution, as is 1. Are there others? Quote Link to comment Share on other sites More sharing options...
0 jasen Posted November 17, 2016 Report Share Posted November 17, 2016 (edited) yes there are still 2 more solution. Spoiler 1458 = 18 * 81 and 1729 = 19 * 91 No more solution, because for numbers bigger than that (addition of digits) * (reverse of addition of digits) will too small for the numbers. Edited November 17, 2016 by jasen Quote Link to comment Share on other sites More sharing options...
Question
bonanova
Inspired by FUZZY's recent puzzle
Suppose abcd is the normal decimal representation of a number: abcd = (1000a)+(100b)+(10c)+c.
The digital sum of a number is defined as a+b+c+d.
Are there numbers for which the product of its digital sum and its reversal is the original number?
For example, the digital sum of 12345 is 15.
Its reversal is 51. 15x51 is 765, 12345 is not equal ti 765,
So 12345 is not a solution.
Hint: 81 is a solution, as is 1.
Are there others?
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