My Math Forum a simple but tricky problem

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 October 31st, 2016, 12:32 PM #1 Senior Member   Joined: Apr 2008 Posts: 192 Thanks: 3 a simple but tricky problem I am stuck in this problem. (ex) Tom sells donuts only in boxes of 7, 13 or 24. To buy 14 donuts, you must order 2 boxes of 7, but you cannot buy exactly 15 donuts since no combinations of boxes contain them. What is the maximum number of donuts that cannot be ordered using combinations of the 3 different size boxes from Tom? This is what I have done. First, I find the least common multiple (LCM) of 7, 13 and 24. The LCM is 2275. Next, I choose and "play around" with the number 2274. It is the biggest number less than the LCM which I think is the answer. But, I don't know how to prove that. Or I could be on the wrong track. Can someone explain how to do it? Thanks a lot.
 October 31st, 2016, 02:58 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 The LCM is 2184. Try to solve the corresponding problem with boxes of 3 or 5, say. When you've done that, try to generalize your answer. Thanks from topsquark
 October 31st, 2016, 03:01 PM #3 Senior Member     Joined: Feb 2010 Posts: 673 Thanks: 125 Well here is a way to solve it (not very mathematical). Make a chart of seven columns and place the integers from 1 up to say 50 (you might have to do more). 1 2 3 4 5 6 7* 8 9 10 11 12 13* 14 15 16 17 18 19 20 21 22 23 24* 25 26* 27 28 29 30 31 32 33 34 35 36 37 38 39* 40 41 42 43 44* 45 46 47 48 49 50* The starred numbers are an attainable number of donuts (for example 50 = 24 + 13 + 13 and 44 = 24 + 13 + 7). All numbers in a column below the stars are also attainable by adding enough 7's. So the largest non-attainable number appears to be 43. Thanks from topsquark
 October 31st, 2016, 06:31 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 In the above table, 37 should be starred (and so 44 doesn't need to be). For those who want to know more about this type of problem, start by reading this article and be prepared to follow links to a considerable depth.

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