 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
Ex
2(Ex)
2(Ex)x
3(2(Ex)x)
3(2(Ex)x)x
4(3(2(Ex)x)x)
4(3(2(Ex)x)x)x
5(4(3(2(Ex)x)x)x)
5(4(3(2(Ex)x)x)x)x
We are told this last equals zero
5(4(3(2(Ex)x)x)x)x = 0
4(3(2(Ex)x)x)x = x/5
3(2(Ex)x)x = (x/5 + x)/4 = 3x/10
2(Ex)x = (3x/10 + x)/3 = 13x/30
Ex = (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 20140428 13:05:56 