1001 Math Problems: other arithmetic puzzles

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Solution: The sum of any two opposite faces of a die is 7. So, excluding the top face, the sum of the numbers on the 3-dice tower is always equal to 7 x 2 x 3 = 42, and it doesn't matter how we arrange the dice. The only thing that affects the total sum is the number showing on the top, which can be any number from 1 to 6. So the minimum sum is 42 + 1 = 43 and the maximum sum is 42 + 6 = 48. For 20 dice, the minimum sum is 7 x 2 x 20 + 1 = 281 and the the maximum sum is 7 x 2 x 20 + 6 = 286. For n dice, the minimum sum is 7 x 2 x n + 1 and the maximum sum is 7 x 2 x n + 6.

The Spilled Juice Problem

Henry had just finished his math homework when he spilled orange juice all over it. Can you help Henry figure out what his homework problems (and solutions!) were?

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Solution:

(1) If one mango and one banana cost $4.70, then two bananas and two mangoes cost twice as much: $9.40.
(2) From there we can use the middle bag to find the cost of a single banana: $13.00 (2 mangoes and 5 bananas) - $9.40 (2 mangoes and 5 bananas) = $3.60 (3 bananas). If 3 bananas cost $3.60, then each banana costs $1.20.
(3) From the first bag then we can see that a single mango costs $3.50 ($4.70-$1.20), and so the bag of 2 mangoes costs $7.00.

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Solution: One way to solve is to combine the 47-cent bag and the 49-cent bag. We then have a 96-cent bag that contains 3 blue candies, 1 green candy, and 2 red candies. If we then subtract the 57-cent bag, we are left with a 39-cent bag that contains 3 blues. So each blue is worth 13 cents. The rest is easy to work out from there.

B = 13, G = 21, R = 18.

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Unique solutions to "More Cryptarithms" Activity:

(1) 91 + 9 = 100 (2) 98 - 89 = 9 (3) 19 + 91 = 110 (4) 101 - 2 = 99

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Solution: There must be a 4 on the back of the 2 chip and a 6 on the back of the 5 chip. (If we allowed larger numbers, there would be one more possible solution -- a 3 on the back of the 2 chip and a 7 on the back of the 5 chip.)