A children's gift store sells gift certificates in denominations of $3 and $5. The store sold 'm' $3 certificates and 'n' $5 certificates worth $93 on a Saturday afternoon. If 'm' and 'n' are natural numbers, how many different values can 'm' take?	Choose the correct choice from the following answers
A) 5  B) 7  C) 6  D) 31  E) 18

Rationale: explanatory answer step 1 : key data 1 . total value of all certificates sold = $ 93 . 2 . certificates sold were in denominations of $ 3 and $ 5 . 3 . both ' m ' and ' n ' are natural numbers . step 2 : approach the value of all certificates sold , 93 is divisible by 3 . so , a maximum of 31 $ 3 certificates and no $ 5 certificates could have been sold . however , the question states that both ' m ' and ' n ' are natural numbers . hence , at least 1 $ 5 certificate should have been sold . let us reduce the number of $ 3 certificates from theoretical maximum count of 31 by say ' x ' and correspondingly increase $ 5 certificates by ' y ' . evidently , 3 x = 5 y because the value of $ 3 certificates reduced should be the same as the value of $ 5 certificates increased . it means that x has to be a multiple of 5 and y has to be a multiple of 3 . or $ 3 certificates reduce in steps of 5 certificates . step 3 : list down possibilities the following combinations are possible 1 . m = 26 , n = 3 2 . m = 21 , n = 6 3 . m = 16 , n = 9 4 . m = 11 , n = 12 5 . m = 6 , n = 15 6 . m = 1 , n = 18 an alternative way to think of the same concept - replacing five $ 3 certificates with three $ 5 certificates leads to no change in the overall value of certficates sold and gives us a new combination each time . we need to see how many such combinations are possible . answer c
Answer: C