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 Making poverty history (Posted on 2006-02-19)
On New Years Day, Penny resolved to make poverty history.

Unfortunately Penny was penniless and asking her relatives for cash seemed a bit passe, so she asked them to contribute goats.

Frankie, Mark, Dolly, Squidly, Ruby and Draco each sent goat vouchers. Each of the vouchers was for a three figure number of goats. Penny was delighted with the response, over 5200 goats in total. Even better, she found that after using 11 of the goats to pay local administrative expenses, each of the beneficiaries would receive exactly 32 goats.

Curiously she noticed that the each of the donations was the product of exactly six prime numbers. If Frankie gave more than Mark, who gave more than Dolly, who gave more than Squidly, who gave more than Ruby, who gave more than Draco, how many did each give?

 No Solution Yet Submitted by goFish Rating: 4.5000 (4 votes)

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 re(4): computer solution | Comment 7 of 9 |
(In reply to re(3): computer solution by Mindrod)

I'm not understanding the 2^5 with regard to the factorization routine. The factorization routine is one that I use whenever I have to factor an integer into primes--not specifically written for this problem.

The first several test divisors the way the loop is written are:

7  11  13  17  19  23  29  31  37  41  43  47  49  53  59  61  67  71  73  77 79  83  89  91  97  101  103  107  109  113  119  121  127  131  133  137  139 143  149  151  157  161  163  167  169  173  179  181  187  191  193  197  199 203  209  211  217  221  223  227  229  233  239  241  247  251  253  257  259 263  269  271  277  281  283  287  289  293  299  301

If the ' 37 line were left out they would be:

7  11  13  17  19  23  29  31  35  37  41  43  47  53  55  59  61  65  67  71 77  79  83  85  89  91  95  101  103  107  109  113  115  119  125  127  131 133  137  139  143  149  151  155  157  161  163  167  173  175  179  181  185 187  191  197  199  203  205  209  211  215  221  223  227  229  233  235  239 245  247  251  253  257  259  263  269  271  275  277  281  283  287  293  295 299  301  305  307  311  317

The first difference is an unnecessary division by 35, but later, more importantly, division by 73 is left out.  Perhaps your comment is related to this specific problem's not needing a prime so high as any of these.  Again, the reason is that I used a generalized factorization subroutine, using a cycle of 30 to avoid trying any multiples of 2, 3 or 5, after the initial trials by those numbers outside the loop.

 Posted by Charlie on 2006-02-19 15:39:45

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