## Children Sharing Money

Al, Bee, Cecil, and Di have $\16,\24,\32$, and $\48$, respectively. Their father proposed that Al and Bee share their wealth equally, and then Bee and Cecil do likewise, and then Cecil and Di. Their mother’s plan is the same except that Di and Cecil begin by sharing equally, then Cecil and Bee, and then Bee and Al. Determine the number of children who end up with more money under their father’s plan than under their mother’s plan.
Source: NCTM Mathematics Teacher, August 2006

Solution
Father’s plan
$(\mathrm{Al}+\mathrm{Bee})/2=(16+24)/2=20$
$(\mathrm{Bee}+\mathrm{Cecil})/2=(20+32)/2=26$
$(\mathrm{Cecil}+\mathrm{Di})/2=(26+48)/2=37$
Al, Bee, and Cecil end up with more money.
Mother’s plan
$(\mathrm{Di}+\mathrm{Cecil})/2=(48+32)/2=40$
$(\mathrm{Cecil}+\mathrm{Bee})/2=(40+24)/2=32$
$(\mathrm{Bee}+\mathrm{Al})/2=(32+16)/2=24$
Only Al ends up with more money.
Two children end up with more money under father’s plan than under mother’s plan.

Answer: $2$