Answer to the April 12, 2004 Problem of the Week

Not Sold At 7-11

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The local convenience store in Winchester sells a gallon of milk for $7 and a loaf of bread for $11.

If you bought 3 milks and 2 breads, you'd spend $43, but there's no way to spend exactly $20.
Starting at $1, how many whole numbers (such as $20) can't equal the "total" of some milks and breads, and what are they?
(It's ok to buy all milk or all bread. Hint: After a certain point, all totals are possible!)

 

Solution to the Problem:

Solution: There are 30 unattainable numbers.

The list includes 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 15, 16, 17, 19, 20, 23, 24, 26, 27, 30, 31, 34, 37, 38, 41, 45, 48, 52, and 59.

Beginning with 60, all numbers can be written as a linear combination of 7 and 11.
60 = 7(7) + 1(11)
61 = 4(7) + 3(11)
62 = 1(7) + 5(11)
63 = 9(7) + 0(11)
64 = 6(7) + 2(11)
65 = 3(7) + 4(11)
66 = 0(7) + 6(11)
Beginning with 67, each remaining number can be written by adding additional sevens to each of the above. For example, 67 can be written as 8(7) + 1(11). Then 68 can be written as 5(7) + 3(11), etc.

James Alarie sent in the following:
The earliest experience with this that I can remember was while eating chicken nuggets at KFC. They came in sets of 6, 9, and 20, and I wondered what numbers of nuggets can't be purchased, whether or not there was a highest forbidden number, and some formula for figuring it out. For there to be a "highest forbidden number," there must be at least two numbers in the set which are relatively prime.
Considering only those two numbers, the answer is X*Y-X-Y, and then you check for combinations using the third, fourth, etc. numbers. For your puzzle, the highest forbidden number is 7*11-7-11=59.

Click here for the Chicken McNuggets Problem