How many pennies?
Hi! It’s great to know that this puzzle got your attention. So let’s begin by revisiting the problem statement:
Solution:
First, let’s assume that the poorest one has x number of pennies. Now, assume that the next child has the number of pennies equal to xy (where y is an integer, as per the given condition). Next, the ratio of the number of pennies that the third child possesses to the number of pennies possessed by the first two must be integer. A simple way of achieving this is to assume that the third child has xyz number of pennies (again, z is an integer). This way, the ratios will be:
xyz:xy = z:1 (integer ratio because z is assumed to be an integer)
xyz:x = yz:1 (integer, because multiplication of two integers will yield another integer).
Applying the same logic throughout, we can assume the fortunes of other children to be:
4th Child: xyzm (m is an integer)
5th Child: xyzmn (n is an integer)
6th Child: xyzmno (o is an integer)
7th Child: xyzmnop (p is an integer)
The sum of all fortunes is 2,879.
Mathematically,
x + xy + xyz + xyzm + xyzmn + xyzmno + xyzmnop = 2879
Taking out x as common:
x (1 + y + yz + yzm + yzmn + yzmno + yzmnop) = 2879
Note that 2879 is a prime number – which means that it can be divided only by itself and 1. In other words, 2879 has only two factors – 1 and 2879.
Therefore, x = 1 and
1 + y + yz + yzm + yzmn + yzmno + yzmnop = 2879
Thus,
y + yz + yzm + yzmn + yzmno + yzmnop = 2879-1
y + yz + yzm + yzmn + yzmno + yzmnop = 2878
Taking y out as common:
y (1 + z + zm + zmn + zmno + zmnop) = 2878 = (2)(1439)
Once again, 1439 is a prime number. Therefore, y = 2
Thus,
1 + z + zm + zmn + zmno + zmnop = 1439
z + zm + zmn + zmno + zmnop = 1438
z (1 + m + mn + mno + mnop) = 1438 = (2)(719)
719 is a prime number. Therefore, z = 2
Solving like this, we get:
m = 2
n = 2
o = 2
Finally,
o (1 + p) = 178
(2)(1 + p) = 178
1 + p = 89
p = 88
Thus, the different fortunes will be:
1st Child: 1 penny
2nd Child: 1×2 = 2 pennies
3rd Child: 1x2x2 = 4 pennies
4th Child: 1x2x2x2 = 8 pennies
5th Child: 1x2x2x2x2 = 16 pennies
6th Child: 1x2x2x2x2x2 = 32 pennies
7th Child: 1x2x2x2x2x2x88 = 2816 pennies
Hope you found this puzzle stimulating! Let us know in the comments section and please give us some love on Facebook 🙂