Question

The remainder of ${51^{51}}$, when divided by 25 is:
$
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. 1}} \\
  {\text{C}}{\text{. 2}} \\
  {\text{D}}{\text{. none of the above}} \\
$

Answer 1

Hint: -To solve this type of question a standard procedure is being followed by which it can be easily solved. Here we start with writing 51 = 50 + 1. Then use binomial expansion to solve further.

Complete step-by-step answer:
We have
${51^{51}}$ now we can write it as ${\left( {50 + 1} \right)^{51}}$
We have studied in the chapter of binomial theorem
${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}{b^1}{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots \ldots { + ^n}{C_n}{a^0}{b^n}$
So applying this binomial expansion on ${\left( {50 + 1} \right)^{51}}$ we get,
${\left( {50 + 1} \right)^{51}}{ = ^{51}}{C_0}{50^{51}}{ + ^{51}}{C_1}{50^{50}}{ + ^{51}}{C_2}{50^{49}} + \ldots \ldots { + ^{51}}{C_{50}}{50^1}{ + ^{51}}{C_{51}}$
Now we will take 50 common from those terms where powers of 50 are available.
${\left( {50 + 1} \right)^{51}} = 50\left( {^{51}{C_0}{{50}^{50}}{ + ^{51}}{C_1}{{50}^{49}}{ + ^{51}}{C_2}{{50}^{48}} + \ldots \ldots { + ^{51}}{C_{50}}} \right) + 1{\text{ }}\left( {{\because ^{51}}{C_{51}} = 1} \right)$
Now let us assume,
\[\left( {^{51}{C_0}{{50}^{50}}{ + ^{51}}{C_1}{{50}^{49}}{ + ^{51}}{C_2}{{50}^{48}} + \ldots \ldots { + ^{51}}{C_{50}}} \right) = m\]
Now we can write it as :
${\left( {50 + 1} \right)^{51}}$ =50m+1
 Now on dividing it by 25 we can clearly see 50m is divisible but 50m+1 when divided by 25 we will get remainder as 1.
Hence option B is the correct option.

Note: -Whenever we get this type of question the key concept of solving is we have to follow the same standard procedure to solve this type of question. We should have remembered binomial expansion formula ${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}{b^1}{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots \ldots { + ^n}{C_n}{a^0}{b^n}$ in many questions this formula is very helpful.