Dani is about to travel on a bus, and she knows she must tender the exact fare. She is not sure what this is, but she knows it is greater than $1 and less than $3. What is the minimum number of coins she must carry to be sure of carrying the correct fare? (Assume that the available coins are 1c, 2c, 5c, 10c, 20c, 50c, $1 and $2)

Combinations:

1 c = 1 c

2 c = 2 c

3 c = 1 c + 2 c

4 c = 2 c + 2 c

5 c = 5 c

6 c = 1 c + 5 c

7 c = 1 c + 2 c

8 c = 1 c + 2 c + 5 c

9 c = 2 + 2 c + 5 c

Prices 1c to 9c she can cover with the minimum number of coins:

1 x 1c , 2 x 2c and 1 x 5c

10 c = 10 c

20 c = 20 c

30 c = 10 c + 20 c

40 c = 20 c + 20 c

50 c = 50 c

60 c = 10 c +50 c

70 c = 20 + 50 c

80 c = 10 c + 20 c + 50 c

90 c = 20 c + 20 c + 50 c

Prices 10c to 90c she can cover with the minimum number of coins:

1 x 10c , 2 x 20c and 1 x 50c

For prices $1 1c to $1 99c she can cover with the minimum number of coins:

1 x $ 1 + 1 x 1c + 2 x 2c + 1 x 5c + 1 x 10c + 2 x 20c + 1 x 50c

For prices $2 to $2 99c she must have:

1 x $ 2 + 1 x 1c + 2 x 2c + 1 x 5c + 1 x 10c + 2 x 20c + 1 x 50c

For prices $1 1 c to $2 99 c the minimum number of coins she must carry:

1 x $1 + 1 x $2 + 1 x 1c + 2 x 2c + 1 x 5c + 1 x 10c + 2 x 20 c + 1 x 50 c

Total 10 coins.

To find the minimum number of coins that Dani must carry to be sure of carrying the correct fare, we need to look for possible combinations of coins that satisfy the given condition.

The minimum fare Dani needs to carry is greater than $1 and less than $3, which means it should lie between 100 cents (equivalent to $1) and 300 cents (equivalent to $3).

Let's go through the denomination of coins available and see which combinations will give us a fare between 100 cents and 300 cents.

1c coin: To get a fare greater than $1, Dani would need at least 101 one-cent coins (101 cents). However, since it is stated that she wants the minimum number of coins, we can exclude the 1c coin since it will not give us the minimum.

2c coin: To get a fare greater than $1, Dani would need at least 51 two-cent coins (102 cents). Again, to minimize the number of coins, we can exclude the 2c coin.

5c coin: To get a fare greater than $1, we would need at least 21 five-cent coins (105 cents). Still not reaching our goal.

10c coin: To get a fare greater than $1, we would need at least 11 ten-cent coins (110 cents). Getting closer, but not there yet.

20c coin: To get a fare greater than $1, we would need at least 6 twenty-cent coins (120 cents). Still not enough.

50c coin: Finally, to get a fare greater than $1, we would need at least 3 fifty-cent coins (150 cents). This satisfies the condition.

Therefore, the minimum number of coins that Dani must carry to be sure of carrying the correct fare is 3, using three 50c coins.