SOLUTION: The 86-digit number 666...666 ( made entirely of sixes) is multiplied by the 85 digit-number 333...333 (made entirely of threes). How many times will the digit 7 appear in the prod

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Question 1140645: The 86-digit number 666...666 ( made entirely of sixes) is multiplied by the 85 digit-number 333...333 (made entirely of threes). How many times will the digit 7 appear in the product?
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


85 times.

Look at the pattern:
66*33 = 2178    2 digits times 2 digits --> 1 7
666*333 = 221778    3 digits times 3 digits --> 2 7's
6666*3333 = 22217778    4 digits times 4 digits --> 3 7's
66666*33333 = 2222177778    5 digits times 5 digits --> 4 7's
...
86 digits time 86 digits --> 85 7's

Here is a demonstration of the result for the case where both numbers are 5 digits; the demonstration for 86-digit numbers will be similar.


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Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
He misread the problem. The multiplier has ONE LESS digit than the multiplicand.
So he should have

66*3 = 198    2 digits times 1 digit --> 0 7
666*33 = 21978    3 digits times 2 digits --> 1 7
6666*333 = 2219778    4 digits times 3 digits --> 2 7's
66666*3333 = 222197778    5 digits times 4 digits --> 3 7's
666666*33333 = 22221977778    6 digits times 5 digits --> 4 7's

So if the pattern continues, then it should be

86 digits times 85 digits --> 84 7's

Edwin
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