The Great Scramble.
After dinner, the five boys of a household happened to find a parcel of sugar-plums. It was quite unexpected loot, and an exciting scramble ensued. You see, Andrew managed to get possession of just two-thirds of the parcel of sugar-plums. Bob at once grabbed three-eighths of these, and Charlie managed to seize three-tenths also. Then, young David dashed upon the scene, and captured all that Andrew had left, except one-seventh, which Edward artfully secured for himself with a cunning trick. Now the fun began in real earnest, for Andrew and Charlie jointly set upon Bob, who stumbled against the table and dropped half of all that he had, which were equally picked up by David and Edward, who had crawled under the table and were waiting. Next, Bob sprang on Charlie from a chair, and upset all the latter's collection on to the floor. Of this prize Andrew got just a quarter, Bob gathered up one-third, David got two-sevenths, while Charlie and Edward divided equally what was left of that stock. They were just thinking the fray was over when David suddenly struck out in two directions at once, upsetting three-quarters of what Bob and Andrew had last acquired. The two latter, with the greatest difficulty, recovered five-eighths of it in equal shares, but the three others each carried off one-fifth of the same. Every Munchkin was now accounted for, and they called a truce, and divided equally amongst them the remainder of the parcel. What is the smallest number of sugar-plums there could have been at the start, and what proportion did each boy obtain?

The answer given was: The smallest number of sugar plums that will fulfil the conditions is 26,880. The five boys obtained respectively: Andrew, 2863; Bob 6,335; Charlie 2,438; David 10,294; Edward 4,950. There is a little trap concealed in the words near the end, “one-fifth of the same,” that seems at first sight to upset the whole account of the affair. But a little thought will shows that the words could only mean “one-fifth of five-eighths, the fraction last mentioned” – that is, one-eighth of the three-quarters that Bob and Andrew had last acquired.

I tried the problem and got the following answer: x was the number of sugar plums. Andrew’s portion was 43/257 of x; Bob’s was 73/809 of x, Charlie’s was 21/562 of x, David’s was 4/15 of x, and Edward’s was 104/795 of x. Now I’m stuck.

One thing I suspect however is that both of you (Dudeney and you) can't be right. You got 5 fractions. Now the task is to find an integer such that any portion of it defined by any of the five fractions is an integer.

The four denominators 257, 809, 562, 795 are mutually prime, i.e., not two of them have common factors, except for 1. This means that the smallest number divisible by all 4 is their product: 92893449270.

If I were you, I would double check your calculations. The fractions should add up to 1, right? In your case it's rather 64319503951/92893449270.