Solution

Given:• \(\frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X\)

To find:• The unit digit of \(\frac{X}{(14!)^7}\)

Approach and Working:Assuming n = 14!, we can rewrite the whole expression as

• X = \(\frac{(n)^{25}-(n)^{11}}{(n)^{11}-(n)^{4}}\)

= \(\frac{(n)^{11}[(n)^{14}-1]}{(n)^4[(n)^{7}-1}\)

= \((n)^7\frac{(n^7+1)(n^7-1)}{(n^7-1)}\)

= \(n^7 (n^7+1)\)

So, if we do \(\frac{X}{(14!)^7}\), we can write it as:

• \(\frac{X}{(14!)^7}\) = \(\frac{n^7 (n^7+1)}{n^7}\) = \(n^7 + 1\)

As n = 14!, n will always end with a 0.

• Therefore, the unit digit of \(n^7 + 1\) = 0 + 1 = 1

Hence, the correct answer is option A.

Answer: A
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