Number of points having position vector ai^+bj^+ck^, where a, b, c ∈ {1, 2, 3, 4, 5} such that 2a + 3b + 5c is divisible by 4 is

4m=2a+3b+5c=2a+(4−1)b+(4+1)c4m=4k+2a+(−1)b+(1)c∴ a = 1, b = even, c = any numbera ≠ 1, b = odd, c = any number∴ Required number = 1 × 2 × 5 × 4 × 3 × 5 = 70