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(Q). How many factors of N = 2⁷ × 3⁷ × 5⁷ × 7⁷ ends with two zeros ? 
Solution:-

In every factor 2² × 5² will be common
∴ required number of factor is same as the number of factors of 2⁵ × 3⁷ × 5⁵ × 7⁷
= 6×8×6×8
= 2304   Answer