Number of Zeroes

Algebra Level pending

How many trailing zeroes are at the end of the product of 13 0 8 130^8 and 54 0 30 540^{30} ?


The answer is 38.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

note that one factor of 10 has one zero at the end, two factors of 10 has two zeroes at the end

130^8 * 540^30 = 10^8 * 13^8 * 10^30 * 54^30

number of zeroes = 8 + 30 = 38

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You should insert the fact that 5 divides neither 13 nor 54. Otherwise, there might be extra zeroes following the product of 2 and 5.

William Nathanael Supriadi - 4 years, 7 months ago

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