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 one
A) 5 B) 7 C) 6 D) 31 E) 18
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, 3x = 5y 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 6
<result>C</result>
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?