Binomial TheoremMedium
Question
The largest $n \in N$, for which $7^{n}$ divides 101!, is :
Options
A.16
B.18
C.15
D.19
Solution
Exponent of 7 in 101 !
$${= \left\lbrack \frac{101}{7} \right\rbrack + \left\lbrack \frac{101}{7^{2}} \right\rbrack + \left\lbrack \frac{101}{7^{3}} \right\rbrack + \ldots\ldots }{= 14 + 2 = 16}$$
Create a free account to view solution
View Solution FreeMore Binomial Theorem Questions
Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^{3}$ in $(1 + x)^{3} + (1 + x)^{4} + (1 + ...If C0, C1, C2,.....Cn denote the binomial coefficients in the expansion of (1 + x)n, then the value of is -...is...If the sum of coefficients in the binomial expansion of (x + y)n is 4096 then greatest coefficient in the expansion is -...If (1 + x + x2)25 = a0 + a1x + a2x2 + ..... + a50. x50 then a0 + a2 + a4 + ..... + a50 is -...