A mathematical problem which shall be linear algebra

Question

I was trying to solve one of the famous H. Dudeney riddles, from Amusement in Maths, and I got stuck with this particular one, which may be as silly as hard. Probably I'm facing it in the wrong way. Here is the whole text:

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, the full details of which I will recount with accuracy, as it forms an interesting puzzle.

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 Edgar artfully secured for himself by a cunning trick. Now the fun began in real earnest, for Andrew and Charlie jointly set upon Bob, who stumbled against the fender and dropped half of all that he had, which were equally picked up by David and Edgar, who had crawled under a 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 Pg 20gathered up one-third, David got two-sevenths, while Charlie and Edgar 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 sugar-plum 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?

answer

The smallest number of sugar plums that will fulfil the conditions is 26,880. The five boys obtained respectively: Andrew, 2,863; Bob, 6,335; Charlie, 2,438; David, 10,294; Edgar, 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 show 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 believe it might just be a tedious calculation but when I try to set down the problem something goes wrong. Any hint? Thank you so much!

Answer 1

I don't know that there is any clever shortcut to solve it. This problem comes from back in times when people loved their word puzzles and were not afraid to do a little arithmetic by hand.

You see, Andrew managed to get possession of just two-thirds of the parcel of sugar-plums.

$A = \frac{2}{3}$

Bob at once grabbed three-eighths of these, and Charlie managed to seize three-tenths also.

$B = \frac{3}{8} A = \frac{1}{4}\,$, $\;C = \frac{3}{10} A = \frac{1}{5}\,$, $\;A = A-B-C = \frac{2}{3}-\frac{1}{4}-\frac{1}{5} = \frac{13}{60}$

Then young David dashed upon the scene, and captured all that Andrew had left, except one-seventh, which Edgar artfully secured for himself by a cunning trick.

$D=\frac{6}{7}A = \frac{13}{70}\,$, $\;E=\frac{1}{7}A = \frac{13}{420}\,$, $\;A = 0\,$

Now the fun began in real earnest, for Andrew and Charlie jointly set upon Bob, who stumbled against the fender and dropped half of all that he had,

$b^{\,'} = \frac{1}{2}B = \frac{1}{2} \frac{1}{4} =\frac{1}{8}\,$, $\;B = B - b^{\,'} = \frac{1}{4}-\frac{1}{8}= \frac{1}{8}$

which were equally picked up by David and Edgar, who had crawled under a table and were waiting.

$D=D+\frac{1}{2} b^{\,'}=\frac{13}{70}+\frac{1}{16} = \frac{139}{560}\,$, $\;E=E+\frac{1}{2}b^{\,'} = \frac{13}{420}+\frac{1}{16}=\frac{157}{1680}\,$

Next, Bob sprang on Charlie from a chair, and upset all the latter's collection on to the floor.

$c^{\,'}=C=\frac{1}{5}\,$, $\;C=0$

Of this prize Andrew got just a quarter, Bob gathered up one-third, David got two-sevenths,

$A=A+\frac{1}{4}c^{\,'} = \frac{1}{20}\,$, $\;B=B+\frac{1}{3}c^{\,'}=\frac{1}{8}+\frac{1}{15}=\frac{23}{120}\,$, $\;D=D+\frac{2}{7}c^{\,'}=\frac{139}{560}+\frac{2}{35}=\frac{171}{560}\,$

while Charlie and Edgar divided equally what was left of that stock.

$\;c^{\,''}=\left(1-\frac{1}{4}-\frac{1}{3}-\frac{2}{7}\right)c^{\,'}=\frac{11}{84}c^{\,'}=\frac{11}{420}\,$, $\;C=C+\frac{1}{2}c^{\,''}=\frac{11}{840}\,$, $\;E=E+\frac{1}{2}c^{\,''}=\frac{157}{1680}+\frac{11}{840}=\frac{179}{1680}$

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.

$b^{\,''}=\frac{3}{4} \frac{1}{3}c^{\,'}=\frac{1}{20}\,$, $\;a^{\,''}=\frac{3}{4}\frac{1}{4}c^{\,'}=\frac{3}{80}\,$, $\;B=B-b^{\,''}=\frac{23}{120}-\frac{1}{20}=\frac{17}{120}\,$, $\;A=A-a^{\,''}=\frac{1}{20}-\frac{3}{80}=\frac{1}{80}\,$

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.

$d^{\,''}=b^{\,''}+a^{\,''}=\frac{1}{20}+\frac{3}{80}=\frac{7}{80}\,$, $\;A=A+\frac{1}{2}\frac{5}{8}d^{\,''}=\frac{1}{80}+\frac{7}{256}=\frac{51}{1280}\,$, $\;B=B+\frac{1}{2}\frac{5}{8}d^{\,''}=\frac{17}{120}+\frac{7}{256}=\frac{649}{3840}\,$, $C=C+\frac{1}{8}d^{\,''}=\frac{11}{840}+\frac{7}{640}=\frac{323}{13440}\,$, $\;D=D+\frac{1}{8}d^{\,''}=\frac{171}{560}+\frac{7}{640}=\frac{1417}{4480}\,$, $\;E=E+\frac{1}{8}d^{\,''}=\frac{179}{1680}+\frac{7}{640}=\frac{1579}{13440}\,$

Every sugar-plum was now accounted for, and they called a truce, and divided equally amongst them the remainder of the parcel.

$z=\frac{1}{5}\frac{1}{3}=\frac{1}{15}\,$:

$$ A=A+z=\frac{51}{1280}+\frac{1}{15}=\frac{409}{3840}=\frac{\color{red}{2863}}{26880} \\[5px] $$ $$ B=B+z=\frac{649}{3840}+\frac{1}{15}=\frac{181}{768}=\frac{\color{red}{6335}}{26880} \\[5px] $$ $$ C=C+z=\frac{323}{13440}+\frac{1}{15}=\frac{1219}{13440}=\frac{\color{red}{2438}}{26880} \\[5px] $$ $$ D=D+z=\frac{1417}{4480}+\frac{1}{15}=\frac{5147}{13440}=\frac{\color{red}{10294}}{26880} \\[5px] $$ $$ E=E+z=\frac{1579}{13440}+\frac{1}{15}=\frac{165}{896}=\frac{\color{red}{4950}}{26880} \\[5px] $$

What is the smallest number of sugar-plums there could have been at the start, and what proportion did each boy obtain?

$\operatorname{lcm}(3840, 768, 13440, 13440, 896) = 26880\,$.