Explanation:

Sol. Ans.(a).

In a multiplication problem, 0 can be produced in two ways. First when a number is multiplied by 10 or second when 5 is multiplied by 2 or any even number.

So to count the number of zeroes in the given multiple let’s break the whole multiplication in to 10, 5 and 2.

We get 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

= 1 × 5 × 10 × (3 × 5) × (2×10) × (5×5) × (3×10) × (7×5) × (4×10) × (9×5) × (5×10) × (11×5) × (6×10)

= 1 × (10 × 10 × 10 × 10 × 10 × 10) × (5 × 5 × 5 × 5 × 5 × 5 × 5 × 5) × (2 × 4 × 6) × (3 × 3 × 7 × 9 × 11)

= 1 × (10 × 10 × 10 × 10 × 10 × 10) × (5 × 5 × 5 × 5 × 5 × 5 × 5 × 5) × {2 × (2 × 2) × (2×3)} × (3 × 3 × 7 × 9 × 11)

Now you can see in above multiplication we have six 10s, eight 5s and four 2s.

Six 10s will give us six 0s while four 2s when combined with four 5s (out of the eight available) will give us additional four 0s.

So in the multiplication six + four = ten 0s will be there.