Puzzles and Problems Archives

December 26, 2010

A math problem

If I multiply 51,249,876 by 3 (thus using all the nine digits once, and once only), I get 153,749,628 (which again contains all the nine digits once). Similarly, if I multiply 16,583,742 by 9 the Pg 16result is 149,253,678, where in each case all the nine digits are used. Now, take 6 as your multiplier and try to arrange the remaining eight digits so as to produce by multiplication a number containing all nine once, and once only. You will find it far from easy, but it can be done.

Select here to see solution
If we multiply 32547891 by 6, we get the product, 195287346. In both cases all the nine digits are used once and once only.
end selection.

Puzzle by Henry Ernest Dudeney, 1917

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