PKN wrote:
If \(\frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X\).
What is the unit digit of\(\frac{X}{(14!)^7}\) ?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
PLEASE
be careful while solving
So nothing you had to do a^2 - b^2 = (a-b) (a+b) & take the common out
\(\frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X\)
take \((14!)^{11}\) common from numerator and \((14!)^{4}\) common from denominator
let 14! be x for sometime
x^11 - x^ 4 * [ x^14 - 1 / x^7 -1 ] = X
x/ x^7 = [ (x^7 - 1) (x^7+ 1) / x^7 -1 ]
x/x^7 = x^7 + 1
put back 14! in the above expression
we will certainly have a zero in the product of 14!
0+1
UD = 1
A
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