### PS2: Problems 1, 2 » Solutions

Apr. 7th, 2010 11:11 am**aranthe**

Here are the solutions to problems 1 and 2. I'll be posting 3/4 as soon as I have it annotated.

#### PS 2: Problem 1

# Initialize configuration variables coefficients = 6, 9, 20 mcnuggets = range( 50, 56 ) terms = len(coefficients) combos = [] # Holds list of combinations. a = 0 # coefficient: 6 b = 0 # coefficient: 9 c = 0 # coefficient: 20 # Loop through the mcnuggets range. for n in mcnuggets: ranges = [] # Holds test ranges. # Create test ranges for the coefficients for this value of n. for i in range(0, terms): limit = int( n / coefficients[i] ) + 1 i_range = range( 0, limit ) ranges.append(i_range) n_combos = [] # Holds all combos for a given n # Loop through the test range. for a in ranges[0]: for b in ranges[1]: for c in ranges[2]: # Check to see if the sum of the terms is equal to n. if (a*coefficients[0]) + (b*coefficients[1]) + (c*coefficients[2]) == n: # If so, create an array of this combination. combo = [ a, b, c ] # Append it to the collection of combos for this n. n_combos.append(combo) #Print combos for this n. print 'For n = ', n print 'Combos:', n_combos

#### PS 2: Problem 2

Theorem:If it is possible to buy x, x+1,…, x+5 sets of McNuggets, for some x, then it is possible to buy any number of McNuggets >= x, given that McNuggets come in 6, 9 and 20 packs.Explain, in English, why this theorem is true.

The key is in the given: The smallest coefficient is 6. Once you find solutions for six consecutive amounts, every amount beyond the last one can be derived by adding 6-packs to one of the consecutive solutions.

In the abstract, the minimum number of consecutive solutions required to insure that every subsequent number has a solution is equal to the smallest coefficient of the terms.