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 Smudge Sludge (Posted on 2004-04-28)
??7
x 3??
?0?3
?1?
?5?
?7??3
The above equation was written on a chalkboard to represent an example of a correctly answered multiplication problem. However, the class prankster went when the teacher wasn't looking and smudged many of the numbers. What did the original problem read?

 Submitted by Gamer Rating: 3.6667 (3 votes) Solution: (Hide) Since the only digit that can end in 3 when multiplied by 7 is 9, that must be the first digit of the bottom multiplicand. Also, since 3 times 7 is 21, the last digit on the row second from the bottom must have a 1 in the ones place. Since the only way ?7 times 3 will give 51 is 17, that must be the next number. ?17 x 3?9 ?0?3 ?1? ?51 ?7??3 Now we can fill in the ?0?3 with 9 times 17 or 53. Also, since 9 times ?17 needs to have an extra 900, the only ?00 times 9 to end in 900 is 1. This means the top multiplicand is 117 and the third line down is 1013 and the fifth line down is 351. 117 x 3?9 1013 ?1? 351 ?7??3 The second column adds up to 7 with no carrying possibility, so the ? in that column must be 1. This means the tens digit in the lower multiplicand must be 1 because it's the only thing that can result in 11? when multiplied by 117, and the ? must also be 7. The bottom three numbers can be added together to get the final solution. 117 x 319 1013 117 351 37283

 Subject Author Date Puzzle Solution With Explanation K Sengupta 2009-01-22 06:02:48 re: Solution incorrect. FK is correct. K Sengupta 2009-01-22 06:00:51 re: Answer JayDeeKay 2009-01-21 13:37:16 Solution incorrect. FK is correct. jduval 2007-09-16 19:57:16 Answer K Sengupta 2007-08-17 12:30:19 No Subject varadarajan 2005-03-03 12:08:38 solution Steve Royer 2004-04-28 21:53:13 Reasoned solution Federico Kereki 2004-04-28 17:58:06 No Subject Saka Devi 2004-04-28 14:11:31

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