Let A=[a_ij],a_ij Z [0,4],1 i,j 2.

Question

Let $A=[{a}_{ij}],{a}_{ij}\in Z\cap [0,4],1\le i,j\le 2$.

The number of matrices A such that the sum of all entries is a prime number $p\in (2,13)$ is \_\_\_\_\_\_\_\_\_\_.

Answer: 204

Accepted Answer

$A=[\begin{gathered} {a}_{11} {a}_{12} \ {a}_{13} {a}_{14} \end{gathered}]$

Such that $\Sigma {a}_{ii}=3,5,7$ or 11

Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.

Then total number of possible matrices

$=4+12+4$

$=20$

For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.

$	herefore $ Total possible matrices $=12+12+12+12+4=52$

For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.

$	herefore $ Total possible matrices $=80$

For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.

$	herefore $ Total number of matrices $=52$

$	herefore $ Total required matrices $=20+52+80+52$

$=204$

Solutions · 1