573! In the trailing Zeros, let us know.
140
The stem asks for the number of trailing zeros in 573!. Each trailing zero needs a factor 10 = 2 × 5, and factors of 2 are always more plentiful, so we count factors of 5.
Use \(\left\lfloor \frac{573}{5} \right\rfloor + \left\lfloor \frac{573}{25} \right\rfloor + \left\lfloor \frac{573}{125} \right\rfloor + \left\lfloor \frac{573}{625} \right\rfloor\).
\(= 114 + 22 + 4 + 0 = 140\).
Hence, 573! has 140 trailing zeros.
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Important Questions from Multiples and Factors
Express 486 as a product of powers of prime factors.
Let p, q, r and s be positive natural numbers having three exact factors including 1 and the number itself. If q > p and both are two-digit numbers, and r > s and both are one-digit numbers, then the value of the expression \(\frac{p-q-1}{r-s}\) is:
Find the greatest three-digit number which is a multiple of 8.
The sum of three consecutive multiples of 7 is 840. The smallest of these multiples is:
If 2160 = 2 a × 3 b × 5 c , then the value of 3 a × 2 -b × 5 -c will be –
