[Guest]

For a positive integer N drawn at random between 2 (base ten) and 2001 (base ten) inclusively, determine the probability that the product of the digits in the base-N representation of 2009 (base ten) is equal to 18 (base ten).

[superprismatic]

The probability is 0.002 = 4/2000.

The possibilities are:

base | representation of 2009 in the base
4 | 133121
223 | 92
1000 | 29
1991 | 1I

[Guest]

1/250, I believe.

Conditions are satisfied for N = 4, 223, 1000, or 1991.

2009₄ = 133121

2009₂₂₃ = 92

2009₁₀₀₀ = 29

2009₁₉₉₁ = 1I (1 18)

I first solved this programatically as follows:

data work.test;


	do n=2 to 2001;			/*For each desired base (base n)*/


		length BASE_N_2009 $100.;

		length digit_c $4.;

		BASE_N_2009 = '';

		remaining = 2009;	/*The number we want to find in our current base*/

		power = 0;			/*The last digit represents the base to the 0th power*/

		dp = 1;				/*Start the digital product at 1*/


		do while (remaining > 0);		/*Figure out the digits in base n one by one*/

			digit = mod(remaining,n**(power+1))/(n**power);	/*Figure out the rightmost unknown digit.*/

			dp = dp*digit;	/*Multiply the digital product by that digit*/

			digit_c = strip(put(digit,$4.));	/*Convert to character*/

			BASE_N_2009 = strip(digit_c) || ' ' || BASE_N_2009;	/*Append the rest of the known digits onto the newly found digit*/

			remaining = remaining - digit*(n**power);	/*Decrement the remaining part of our number*/

			power + 1;		/*Move onto the next digit*/

		end;

		drop digit digit_c power remaining;

		output;	

	end;


run;


data work.test;

	set work.test;

	where dp=18;

run;


proc print data=work.test noobs

run;

Results:

  The SAS System        15:31 Tuesday, March 16, 2010   2


                                      n    BASE_N_2009    dp


                                      4    1 3 3 1 2 1    18

                                    223    9 2            18

                                   1000    2 9            18

                                   1991    1 18           18

Basically, I brute forced the solutions for anything with 4 or more digits in base N (up through base 12), stopping when a digit was anything other than a factor of 18.

Trying all the possibilities (only 6 in total) where two digits could have a product of 18 and could be equal to 2009₁₀ in an integer base yielded the last three results.

Now to show that there are no desirable solutions if 2009₁₀ has 3 digits in base N.

Suppose that such a solution, when 2009 has 3 digits in base N, exists. Then there must be a solution for AN² + BN + C = 2009, where A, B, C, and N are positive integers, and A, B, and C are all factors of 18.

Let the factors of AN² + BN + C = (RN+K₁)(SN+K₂) for positive integers R, S, N, K₁, and K₂, and RS = A, and K₁*K₂ = C, where C is a factor of 18. Thus, K₁ and K₂ are also factors of 18. In the same way, as RS = A, R and S must be factors of 18. This means that the factors of AN² + BN + C = 2009 are also factors of 18. This cannot be true, as GCF(18,2009) = 1, so 2009 cannot have 3 digits in base N under our conditions.

Edited March 16, 2010 by Chuck