A children's gift store sells gift certificates in denominations of $3 and $5.
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?
Answer/Solution
6
Steps/Work
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 C
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 C