• Myra took a certain number of eggs to the market and sold some of them.
• The next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.
• On the third day the new remainder was tripled, and she sold the same number as before.
• On the fourth day the remainder was quadrupled, and her sales the same as before.
• On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock.
• Each of the number of eggs that Myra took to the market and the number of eggs that she sold on a given day is a positive integer.

What is the minimum number of eggs she could have taken to market the first day, and how many did she sell daily?

E=eggs at beginning, x=number sold each day
E
E-x
2(E-x)
2(E-x)-x
3(2(E-x)-x)
3(2(E-x)-x)-x
4(3(2(E-x)-x)-x)
4(3(2(E-x)-x)-x)-x
5(4(3(2(E-x)-x)-x)-x)
5(4(3(2(E-x)-x)-x)-x)-x
We are told this last equals zero
5(4(3(2(E-x)-x)-x)-x)-x = 0
4(3(2(E-x)-x)-x)-x = x/5
3(2(E-x)-x)-x = (x/5 + x)/4 = 3x/10
2(E-x)-x = (3x/10 + x)/3 = 13x/30
E-x = (13x/30 + x)/2 = 43x/60
E = 43x/60 + x = 103x/60
60E = 103x
smallest positive integer solution is E=103, x=60

Checking
103 - 60 = 43
43*2 = 86
86 - 60 = 26
26*3 = 78
78 - 60 = 18
18*4 = 72
72 - 60 = 12
12*5 = 60
60 - 60 = 0
0 Eggs left

Posted by Jer on 2014-04-28 13:05:56