Bunuel
There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?
I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
II. If the students are distributed into groups that contain 3 students each, no students are left without a group
III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
This problem lacks any clear solutions. Here is mine:
For some constants k and c:
We know N = 4Ak + 3 and N = A/3 * c.
A/3 must be an integer.
I. Is it possible for N = (A+1)k + 2?First let's find possible values for A:
4Ak + 3 = A/3 * c.
12Ak + 9 = Ac
k is an arbitrary constant, so we can say 12Ak = Ak.
Ak + 9 = Ac
9 = A(c-k)
Again, c and k are arbitrary constants, so we can say c-k = k
9 = Ak
A must be a factor of 9. Given that A/3 is an integer, A can be 3 or 9.
Case 1, A = 3:
Can N = 4c + 2?
4Ak + 3 = 4c + 2
If k is arbitrary, then Ak = k
4k + 3 = 4c + 2
4k + 1 = 4c
This is never true for integer values of c and k
rejectedCase 2, A = 9:
Can N = 10c + 2?
From N = 4Ak + 3, N = 36k + 3
36k + 3 = 10c + 2
36k + 1 = 10c
2*(18k) + 1 = 2*(5c)
as k and c are arbitrary, k = 18k and c = 5c
2k + 1 = 2c
This is never true for integer values of c and k
rejected
Answer to I: No
II. Is N = 3k?
N = 4Ak + 3
A/3 is an integer
N = 4 * 3 * A/3 * k + 3
N = 3(4 * A/3 * k + 1)
k is arbitrary, so 4 * A/3 * k = k
N = 3(k+1)
k is arbitrary, so k+1 = k
N = 3k
Answer to II: Yes
III. Is N = 12k + 9?
N = 4Ak + 3
A/3 is an integer
N = 4 * 3 * A/3 * k + 3
N = 12 * A/3 * k + 3
k is arbitrary, so A/3 * k = k
N = 12k + 3
12k + 3 can never be 12k + 9, for some integer value of k
Answer to III: No
Final Answer: B) II onlyExplanation using Modular Arithmetic:N = 3 (mod 4A)
N = 0 (mod A/3)
A/3 is an integer
I. Can N = 2 (mod A+1)?4Ak + 3 = 0 (mod A/3)
12Ak + 9 = 0 (mod A)
9 = 0 (mod A)
As A/3 is an integer, A can be 3 or 9.
Case 1, A = 3
N = 2 (mod 4)?
N = 3 (mod 4A) -> N = 3 (mod 4)
rejectedCase 2, A = 9
N = 2 (mod 10)?
36k + 3 = 2 (mod 10)
6k + 3 = 2 (mod 10)
6k = -1 (mod 10)
6k = 9 (mod 10)
2k = 3 (mod 10)
2 divides 10 but not 3, so k DNE
rejectedAnswer to I: NoII. N = 0 (mod 3)?N = 3 (mod 4A)
A/3 is an integer
N = 3 (mod 4*3*A/3)
N = 3 (mod 3)
3 = 0 (mod 3)
N = 0 (mod 3)
Answer to II: YesIII. Is N = 9 (mod 12)?N = 3 (mod 4A)
A/3 is an integer
N = 3 (mod 4*3*A/3)
N = 3 (mod 12 * A/3)
N = 3 (mod 12)
Answer to III: NoFinal Answer: B) II only