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| Remainder With 3’s And 7’s (Posted on 2008-11-02) |
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P is a positive decimal (base 10) integer consisting entirely of the digit 3, and Q is a positive decimal integer consisting entirely of the digit 7. In the base-10 expansion of P*Q, the digit 3 is repeated precisely three times and the digit 7 is repeated precisely seven times. The product P*Q may consist of other digits besides 3 and 7.
Given that N is the minimum value of P*Q, determine the remainder when N is divided by 37.
Note: Try to derive a non computer assisted method, although computer programs/spreadsheet solutions are welcome.
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Submitted by K Sengupta
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Rating: 4.5000 (2 votes)
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Solution:
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(Hide)
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Using analytical methodologies in this location, Dej Mar has shown that:
N = 21*R(22)*R(25), where R(p) is the pth repunit.
Now, using the divisibility rule for 37, it is trivial to note that any number of the form R(3x+1), would always leave 1 as the remainder when divided by 37.
Thus, N (mod 37) = 21*1*1 (mod 37) = 21, and accordingly, the required remainder is 21.
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