Question

The greatest number which divides ${25}^n-24n-1$ for all $n\in N$ is

• A. $26$
• B. $578$
• C. $27$
• D. $576$

Answer 1

Answer: (D)

$576$

Consider given the expression,

$\Rightarrow {25}^n-24n-1\forall n\in N$

Now,

${25}^n={(1+24)}^n$

${(1+24)}^n{=}^n{C}_0{+}^n{C}_1.24{+}^n{C}_2{.24}^2{+}^n{C}_3{.24}^3{.............}^n{C}_n{.24}^n$

${25}^n=1+n.24{+}^n{C}_2{.24}^2{+}^n{C}_3{.24}^3{.............}^n{C}_n{.24}^n$

${25}^n-24n-1{=}^n{C}_2{.24}^2{+}^n{C}_3{.24}^3{.............}^n{C}_n{.24}^n$

${25}^n-24n-1={24}^2{[}^n{C}_2{+}^n{C}_3{.24.............}^n{C}_n{.24}^{n-2}$

${25}^n-24n-1=576{[}^n{C}_2{+}^n{C}_3{.24.............}^n{C}_n{.24}^{n-2}$

Hence, ${25}^n-24n-1$ is divisible by 576.

Hence, this is the answer.