Explanation:We will use three key concepts to solve this question;
1. When we divide \(25!\) by \(10^7\), the last 7 digits would be the remainderex: if we divide \(5!\) by 100, the last 2 digits would be the remainder
\(5! = (5)(4)(3)(2)(1) = 120\)
Remainder =
202. In order to find the maximum power of any number (say \(a\)) in \(N!\) (say), divide \(N!\) by the increasing power of \(a\) till the numerator remains greater than the denominatorex: to find number of 2's in \(4!\):
\(\frac{4}{2^1} + \frac{4}{2^2}\)
\(2 + 1 = 3\)
Therefore, number of 2's in \(4!\) is 3
3. Cyclicity of 2, 3, 7 and 8 is 4 whereas cyclity of 4 and 9 is 2i.e. the units digit of 2, 3, 7, and 8 repeat every 4 powers and units digit of 4 and 9 repeat every 2 powers
Coming back to the Question:
Find the number of 2's, 3's, 5's, 7's, 11's, 13's, 17's, 19's and 23's in \(25!\) by using the second concept.
Number of 2's = 22
Number of 3's = 10
Number of 5's = 6
Number of 7's = 3
Number of 11's = 2
Number of 13's = 1
Number of 17's = 1
Number of 19's = 1
Number of 23's = 1
i.e. \(25! = (2^{22})(3^{10})(5^6)(7^3)(11^2)(13)(17)(19)(23)\)
\(25! = (2^{16})(3^{10})(7^3)(11^2)(13)(17)(19)(23)(10^6)\)
Now, Units digit of:
\((2^{16})\) is 6
\((3^{10})\) is 9
\((7^3)\) is 3
\((11^2)\) is 1
\((13)\) is 3
\((17)\) is 7
\((19)\) is 9
\((23)\) is 3
So, \(25! = (ab6)(cd9)(ef3)(gh1)(ij3)(kl7)(mn9)(op3)(10^6) = abcdefghijklmnop 4(10^6)\)
Col. A: \(4(10^6)\)
Col. B: \(2(10^6)\)
Hence, option A
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/