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number theory
- Thread starter benjiboo
- Start date
We have the equation for the total take.
\(\displaystyle 1.8a+.75c=90\)
Which we can write as \(\displaystyle 180a+75c=9000\)
This reduces to \(\displaystyle 12a+5c=600\)
Solving for a we get \(\displaystyle a=\frac{600-5c}{12}\)
'a' must be an integer, as well as c.
We are also told that a>c.
You can try using the Euclidean algorithm to solve the Diophantine equation, and write 600 as a linear combination of 12 and 5.
But, perhaps use trial and error. Enter in c values, get an integer result for a. But, make sure a is greater than c. That is a restriction.
\(\displaystyle 1.8a+.75c=90\)
Which we can write as \(\displaystyle 180a+75c=9000\)
This reduces to \(\displaystyle 12a+5c=600\)
Solving for a we get \(\displaystyle a=\frac{600-5c}{12}\)
'a' must be an integer, as well as c.
We are also told that a>c.
You can try using the Euclidean algorithm to solve the Diophantine equation, and write 600 as a linear combination of 12 and 5.
But, perhaps use trial and error. Enter in c values, get an integer result for a. But, make sure a is greater than c. That is a restriction.
benjiboo said:A cinema charges £1.80 for adults and 75p for children at its matinee screenings during the holidays. The takings for one such matinee amounted to £90. More adults than children attended! How many people attended?
One approach
Let A = Adults and C = children
1--180 A + 75C = 9000 or 12A + 5C = 600
2--Dividing through by the lowest coefficient yields 2A + 2A/5 = 120
3--2A/5 must be an integer k making A = 5k/2
4--Subbing back into (1) yields C = 120 - 6k
5--The first value of k that yields D greater than C is k = 16 where A = 40 and C = 24
Only values of k = 16 and 18 provide solutions where A exceeds Cwith C > 0.
(Thanks to Subhotosh Khan as my hands were quicker than the eye, or the brain)
D
Deleted member 4993
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Only 3 values of k are possible:
k = 16 ? A = 40 & C = 24
k = 18 ? A = 45 & C = 12
k = 20 ? A = 50 & C = 0
k = 16 ? A = 40 & C = 24
k = 18 ? A = 45 & C = 12
k = 20 ? A = 50 & C = 0