Here is an interesting question that combines linear equations and properties of numbers to get an answer.
Question
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?
A. 5
B. 7
C. 6
D. 31
E. 18
Correct Answer : Choice C. 6 different values
Explanatory Answer
Question
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?
A. 5
B. 7
C. 6
D. 31
E. 18
Correct Answer : Choice C. 6 different values
Explanatory Answer
Key data :
1. Total value of all certificates sold = $93.
2. Certificates sold were in denominations of $3 and $5.
2. Certificates sold were in denominations of $3 and $5.
3. Both ‘m’ and ‘n’ are natural numbers.
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.
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.
If we reduce the number of $3 certificates from the maximum 31 that is possible by say ‘x’ and correspondingly increase $5 certificates by ‘y’, then 3x = 5y as 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.
So, 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
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
Think about it another way. Replacing five $3 certificates with three $5 certificated leads to no change in the overall value of certficates sold and gives us a new combination every time. We need to see how many times this can be done.
Queries, answers, comments welcome