2 9 7 x  1 8
More Strange Multiplication
= 5346

     Here are some more multiplication puzzles for you to solve.

     BUT, this time, unlike the page Nine-Digits-&-Equal-Products, where we use all nine digits to make two identical products, we only have one multiplication here. So the nine digits make the two factors and the product at the same time.

     Here is an example:     297 × 18 = 5346

     As far as the two factors and the product are concerned, only the digits from 1 to 9 show up--once and only once!

     Some of the problems below are just like the example above, and some are imposters. Fill in the blanks with the products of ONLY those problems that are like the example. (Leave the others blank.)

  • 157 × 28 = _ _ _ _ _

  • 159 × 27 = _ _ _ _ _

  • 186 × 39 = _ _ _ _ _

  • 198 × 27 = _ _ _ _ _

  • 276 × 13 = _ _ _ _ _

  • 159 × 48 = _ _ _ _ _

     In the last four problems, the digits of the factors are missing and replaced by *** (stars). Replace the stars with the correct digits in order to have the product shown.

  • * * * * × * = 7852

  • * * * * × * = 6952

  • * * * × * * = 5796     [Hint: the 2-place factor is less than 20.]

  • * * * × * * = 5796     [Hint: this one has the same product but different factors. Here the 2-place factor is greater than 40.]


Source for idea: H. E. Dudeney, Amusements in Mathematics. Dover, 1958. pp. 15 & 156.
tt(1/23/77)

Update: 7/23/01

     Recently I have met another recreational mathematician likewise interested in the topic shown on this page. His name is G. L. Honaker, Jr. and his webpage is Prime Curios. He has an extension of the theme above that uses all the ten (10) digits. He asks:

     Can you find the two factors that yield the product of 28651, using the remaining five (5) digits?

     G.L. also has his own special variation on the "9 digits" style, too. Here are two products -- 8614 and 5986. Find a way to use the five remaining digits in each case to form three factors, producing those numbers. It's easier than it looks.

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