Question

B = { blb is a factor of 6, b EU}​

Answer 1

Answer:

If v2(a)=r then r≤v2(b)≤6. Of course, v2(a)∈{0,1,2,3,4,5,6}. If v2(a)=0 there are 7 possibilities for v2(b). If v2(a)=1 there are 6 possibilities for v2(b), and so on. Thus, considering only powers of 2, we get

7+6+5+4+3+2+1=7×82=28

possible pairs.

The same calculation works for the powers of 3.

As we can sort out powers of 2,3 independently, we get

28×28=784

Answer 2

Answer:

If v2(a)=r then r≤v2(b)≤6. Of course, v2(a)∈{0,1,2,3,4,5,6}. If v2(a)=0 there are 7 possibilities for v2(b). If v2(a)=1 there are 6 possibilities for v2(b), and so on. Thus, considering only powers of 2, we get

7+6+5+4+3+2+1=7×82=28

possible pairs.

The same calculation works for the powers of 3.

As we can sort out powers of 2,3 independently, we get

28×28=784