243! In the trailing Zeros, let us know.
58
The number of trailing zeros in n! equals the number of times 5 appears as a factor, found by \(\left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \dots\).
\(\left\lfloor \frac{243}{5} \right\rfloor = 48\), \(\left\lfloor \frac{243}{25} \right\rfloor = 9\), \(\left\lfloor \frac{243}{125} \right\rfloor = 1\), and \(\frac{243}{625}\) is less than 1.
Total = 48 + 9 + 1 = 58.
Hence, 243! has 58 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 –
