A one-dollar coin should have a mass of $7.0 \mathrm{~g}$. Each individual coin may be lighter or heavier by as much as $2.14 \%$. Joshua has a number of these coins and determines that they have a total mass of $1 \mathrm{~kg}$. What is the difference between the greatest possible number and the least possible number of these coins that he could have?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
The maximum possible mass of a given coin is $7 \times(1+0.0214)=7 \times 1.0214=7.1498 \mathrm{~g}$. The minimum possible mass of a given coin is $7 \times(1-0.0214)=7 \times 0.9786=6.8502 \mathrm{~g}$. What are the possible numbers of coins that could make up $1000 \mathrm{~g}$ ?

To find the largest number of coins, we want the coins to be as light as possible. If all of the coins were as light as possible, we would have $\frac{1000}{6.8502} \approx 145.98$ coins. Now, we cannot have a non-integer number of coins. This means that we must have at most 145 coins. (If we had 146 coins, the total mass would have to be at least $146 \times 6.8502=1000.1292 \mathrm{~g}$, which is too heavy.) Practically, we can get 145 coins to have a total mass of $1000 \mathrm{~g}$ by taking 145 coins at the minimum possible mass and making each slightly heavier.

To find the smallest number of coins, we want the coins to be as heavy as possible. If all of the coins were as heavy as possible, we would have $\frac{1000}{7.1498} \approx 139.86$ coins. Again, we cannot have a non-integer number of coins. This means that we must have at least 140 coins. (If we had 139 coins, the total mass would be at most $139 \times 7.1498=993.8222 \mathrm{~g}$, which is too light.) Therefore, the difference between the largest possible number and smallest possible number of coins is $145-140=5$.

ANSWER: (B)