Let A be a matrix of order 2 × 2, whose entries are from the set 0, 1, 2, 3, 4, 5. If the sum of all the entries of A is a prime number p, 2 < p < 8, then the number of such matrices A is ___________.

Solution

Maths sir · 3 steps

Four entries adding to a fixed prime is a stars-and-bars count, done once for each of 3, 5 and 7.

1. Which primes. The only primes with 2 < p < 8 are 3, 5 and 7. Call the four entries a, b, c, d, each a whole number from 0 to 5, and count the solutions of a + b + c + d = p for each prime.

   #{a+b+c+d = n, a,b,c,d ≥ 0} = C(n+3, 3)

2. Remove entries above 5. No entry can be more than 5. For p = 3 and p = 5 no entry can reach 6 anyway. For p = 7, an entry of 6 or 7 is possible. Take a ≥ 6 and write a = 6 + a', so a' + b + c + d = 1, which has C(4, 3) = 4 solutions. Any of the 4 entries could be the big one, so remove 4 × 4 = 16.

   Prime p	All solutions C(p+3, 3)	Entry above 5	Valid matrices
   3	C(6, 3) = 20	0	20
   5	C(8, 3) = 56	0	56
   7	C(10, 3) = 120	16	104

3. So. Each solution (a, b, c, d) fills the four positions of A in exactly one way. The number of matrices is 20 + 56 + 104 = 180.

The common slip: 196. Forgetting the upper limit of 5 for p = 7 keeps the 16 cases with an entry of 6 or 7, giving 20 + 56 + 120 = 196 instead of 180.

The idea to keep. Entries with a fixed sum: use stars and bars, C(n+r-1, r-1). Then check the upper limit on each entry and remove the cases that cross it.

Answer from the NTA final answer key. Solution by PrepWiser.