Using pennies, nickels, dimes or quarters, determine which kind of coin is in each circle in the answer Colum. First clue: The total value of the four coins is an odd number.

Second clue: The sum of the value of the middle 2 coins is 11 cents.
Third clue: Each coin is worth the same as or more than the coin to its left (coin C is worth the same as or more than the coin B).
The order for the circles is circle A, B, C, D. Please explain how you got your answer.

a <= b <= c <= d

b+c = 11
so, since b<=c, b=1 c=10
so, b+c is odd, a+d is even, and
since b=1, a=1 since a <= b
if a+d is even, is odd and the coins are

1 1 10 25

since 10 <= d

To solve this problem, let's follow the given clues and determine the type of coin in each circle:

First, let's consider the first clue, which states that the total value of the four coins is an odd number. We know that pennies are the only coin worth one cent, and quarters are the only coin worth 25 cents. Since an odd number can only be obtained by adding an odd number of 1-cent coins or an odd number of 25-cent coins (or both), we have two possibilities:
- 1 penny, 1 penny, 1 penny, 21 cents (3 odd number of 1-cent coins)
- 1 quarter, 1 quarter, 1 quarter, 9 cents (1 odd number of 25-cent coins and 1 odd number of 1-cent coins)

Next, let's move on to the second clue, which states that the sum of the value of the middle 2 coins is 11 cents.
- In the first possibility, where we have 3 pennies and 1 quarter, the sum of the middle two coins (1 penny + 1 penny) is 2 cents, not 11 cents. So, this possibility is not valid.
- In the second possibility, where we have 1 quarter and 1 penny, the sum of the middle two coins (1 quarter + 1 penny) is 26 cents, again not 11 cents. So, this possibility is also not valid.

Now, let's consider the third clue, which states that each coin is worth the same as or more than the coin to its left.
- In both possibilities, the value of the coins from left to right is decreasing. So, none of the possibilities satisfy this clue.

Based on our analysis, we can conclude that there is no valid solution that satisfies all three clues.