Answer:
A. $301
B. $721
Step-by-step explanation:
Let $x be the amount of money they raised.
Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then
[tex]x=2q_1+1[/tex]
Polly tried to put the $1 bills into three equal piles and found one left over at the end, then
[tex]x=3q_2+1[/tex]
Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then
[tex]x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1[/tex]
Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then
[tex]x=7q_6[/tex]
This means [tex]x-1[/tex] is divisible by 2, 3, 4, 5 and 6 without remainder, so
[tex]x-1=2\cdot 3\cdot 2\cdot 5n=60n[/tex]
Hence,
[tex]x=60n+1, \ n\in N[/tex]
The smallest amount of money they could have raised is $301, because
[tex]x=60\cdot 5+1=301[/tex] is divisible by 7.
Now, the number [tex]x=60n+1[/tex] should be divisible by 7 and must be greater than 500.
So,
[tex]60n+1>500\\ \\60n>499\\ \\n>8[/tex]
When n = 9,
[tex]x=60\cdot 9+1=541[/tex] is not divisible by 7.
When n = 10,
[tex]x=60\cdot 10+1=601[/tex] is not divisible by 7.
When n = 11,
[tex]x=60\cdot 11+1=661[/tex] is not divisible by 7.
When n = 12,
[tex]x=60\cdot 12+1=721[/tex] is divisible by 7.
B. The least amount of money they could have raised is $721
