The town of Hamlet has 3 people for each horse, 4 sheep for each cow, and 3 ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
Answer Choices:
A. 41
B. 47
C. 59
D. 61
E. 66
+6248 you don't give any choices...
total = p + h + s + c + d = 3h + h + 4c + c + 3(3h) = 13h + 5c
here's the first 100 possible totals
18, 23, 28, 31, 33, 36, 38, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56,
57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 72, 74, 75, 76, 77,
79, 80, 82, 83, 84, 85, 87, 88, 89, 90, 92, 93, 95, 96, 97, 98, 100,
101, 102, 103, 105, 106, 108, 109, 110, 111, 113, 114, 115, 116, 118,
119, 121, 122, 123, 124, 126, 127, 128, 129, 131, 132, 134, 135, 136,
137, 139, 140, 141, 142, 144, 145, 147, 149, 150, 152, 154, 155, 157,
160, 162, 165, 167, 170, 175, 180
+6248 you just plug in h=1,10, c=1,10
and sort them
I guess what I'd do is solve 13h+5c = (41,47,59,61,66)
and remember that h and c have to be non-negative integers
\(13h+5c = 41\\ c(h) = \dfrac{41-13h}{5}\\ c(2) = 3\\ \text{so 41 is a legit choice}\)
\(13h+5c=47\\ c(h)=\dfrac{47-13h}{5}\\ \text{there are no integer solutions to this equation}\\ \text{so this must the the non legit total}\)
checking for thoroughness
\(13\cdot 3 + 5\cdot 4 = 59\)
\(13\cdot 2 + 5 \cdot 7 = 61\)
\(13 \cdot 2 + 5 \cdot 8 = 66\)
For example, if I were taking a test, what would be an easier way to get my answer choice, which is B, instead of writing out all the possible numbers?
+118608 The town of Hamlet has 3 people for each horse, 4 sheep for each cow, and 3 ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
Answer Choices:
A. 41
B. 47
C. 59
D. 61
E. 66
1Horse, 3People, 9ducks = some multiple of 13
1Cow, 4 sheep = some multiple of 5
So just like Rom said the answer will be such some multiple of 13H+5C
Now the multiples of 13 are 0, 13, 26, 39, 42, 55, 68 and 68 is bigger than any choice so I do not need to go further
5C will end in 0 or 5 so I can add 0 or 5 to the last digit of the multiples of 13 above
41=26+15
47= ??? to get an answer of 47 the multiple of 13 would have to end in 7 or 2 and none do so it is not a possible answer.
