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Remainder With 3ís And 7ís (Posted on 2008-11-02) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution computer solution | Comment 2 of 3 |

N will be the product of 3*7 times two repunit integers (integers consisting of repeated 1's in decimal notation).  In general, the size of the number N will increase as the total of 1's in the two repunits' representations.

The following UBASIC program goes through increasing total number of 1's in the repunits:

   10      for T=2 to 1000
   20       for A=1 to int(T/2)
   30        B=T-A
   40        R1=int((10^A-1)/9)
   50        R2=int((10^B-1)/9)
   60        Prod=21*R1*R2
   70        S=cutspc(str(Prod))
   80        Ct3=0:Ct7=0
   90        for I=1 to len(S)
  100          if mid(S,I,1)="3" then inc Ct3
  110          if mid(S,I,1)="7" then inc Ct7
  120        next
  130        if Ct3=3 and Ct7=7 then
  131         :print A,R1:print B,R2
  132         :print Prod:print Prod@37:print T:print log(Prod)/log(10)
  140       next
  150      next

It finds only

22      1111111111111111111111
25      1111111111111111111111111

meaning that the smaller repunit has 22 1's and the larger has 25, so P and Q can consist of 22 3's and 25 7's, respectively or 25 3's and 22 7's.

The resulting N is 25925925925925925925923330740740740740740740741.

  Posted by Charlie on 2008-11-02 23:43:33
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