Question

PI-3.

Answer 1

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

$x=2q_1+1$

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

$x=3q_2+1$

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

$x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1$

Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

$x=7q_6$

This means $x-1$ is divisible by 2, 3, 4, 5 and 6 without remainder, so

$x-1=2\cdot 3\cdot 2\cdot 5n=60n$

Hence,

$x=60n+1, \ n\in N$

The smallest amount of money they could have raised is $301, because

$x=60\cdot 5+1=301$ is divisible by 7.

Now, the number $x=60n+1$ should be divisible by 7 and must be greater than 500.

So,

$60n+1>500\\ \\60n>499\\ \\n>8$

When n = 9,

$x=60\cdot 9+1=541$ is not divisible by 7.

When n = 10,

$x=60\cdot 10+1=601$ is not divisible by 7.

When n = 11,

$x=60\cdot 11+1=661$ is not divisible by 7.

When n = 12,

$x=60\cdot 12+1=721$ is divisible by 7.

B. The least amount of money they could have raised is $721

Answer 2

Answer:

A:301 and b: 721