Question

Find the last two digits of the number \[{{14}^{{{14}^{14}}}}\].

Answer 1

Hint: Use binomial expansion to simplify the given number and write it as \[{{14}^{{{14}^{14}}}}={{196}^{\dfrac{{{14}^{14}}}{2}}}={{\left( 200-4 \right)}^{{{7.14}^{13}}}}\]. Consider the possible last two digits of powers of 4 to find a pattern to find the last two digits of the given number.

Complete step-by-step answer:

We have to find the last two digits of the number \[{{14}^{{{14}^{14}}}}\]. We will try to simplify the given number using binomial expansion.
We can write \[{{14}^{{{14}^{14}}}}\] as \[{{\left( {{\left( 14 \right)}^{2}} \right)}^{\dfrac{{{14}^{14}}}{2}}}\] and further as \[{{196}^{\dfrac{{{14}^{14}}}{2}}}\] because \[{{14}^{2}}=196\].

Thus, we have \[{{14}^{{{14}^{14}}}}={{196}^{\dfrac{{{14}^{14}}}{2}}}\].

We can write 196 as \[196=200-4\].

Thus, we have \[{{14}^{{{14}^{14}}}}={{196}^{\dfrac{{{14}^{14}}}{2}}}={{\left( 200-4 \right)}^{{{14}^{14}}}}\].

We can rewrite the above expression as \[{{14}^{{{14}^{14}}}}={{196}^{\dfrac{{{14}^{14}}}{2}}}={{\left( 200-4 \right)}^{\dfrac{14\times {{14}^{13}}}{2}}}={{\left( 200-4 \right)}^{7\times {{14}^{13}}}}\].
Thus, the last two digits of \[{{14}^{{{14}^{14}}}}\] will be the same as the last two digits of \[{{\left( 200-4 \right)}^{7\times {{14}^{13}}}}\].

We know the Binomial Theorem which states that \[{{\left( x+y \right)}^{n}}=\sum\limits_{i=1}^{n}{{}^{n}{{C}_{i}}{{x}^{i}}{{y}^{n-i}}}\].

We observe that in the expansion of \[{{\left( 200-4 \right)}^{7\times {{14}^{13}}}}\], the last two digits of the power of 200 will be 00. Thus, we will only consider the last two digits of \[{{\left( -4 \right)}^{7\times {{14}^{13}}}}={{4}^{7\times {{14}^{13}}}}\].

Now, we observe that the last two digits of even powers of 4 has the following 5 terms occurring periodically \[\left\{ 16,56,96,36,76 \right\}\] as last two digits of \[{{4}^{2}}\] is 16, \[{{4}^{4}}\] is 56, \[{{4}^{6}}\] is 96, \[{{4}^{8}}\] is 36 and \[{{4}^{10}}\] is 76. The last two terms of \[{{4}^{12}},{{4}^{14}},{{4}^{16}},...\] is repeated again.

Now, we observe that \[{{4}^{7\times {{14}^{13}}}}\] has an even exponent of 4. We need to find the last digit of this exponent and match it with any one of the five cases above.
So, we will now find the last digit of \[7\times {{14}^{13}}\], which is the same as the last digit of \[7\times {{\left( 10+4 \right)}^{13}}\].

We observe that in the expansion of \[{{\left( 10+4 \right)}^{13}}\], the last digit of the power of 10 will be 0. Thus, we will only consider the last two digits of \[{{4}^{13}}\].

Thus, the last digit of \[7\times {{14}^{13}}\] is the same as the last digit of \[7\times {{4}^{13}}\], which is the same as the last digit of \[7\times 4\times {{4}^{12}}=28\times {{4}^{12}}\].

We know that the last digit of \[{{4}^{12}}\] is 6. Thus, the last digit of \[28\times {{4}^{12}}\] is the same as the last digit of \[8\times 6=48\], which is 8.

Thus, the last digit of \[7\times {{14}^{13}}\] is 8.

As we know that the last two digits of \[{{4}^{7\times {{14}^{13}}}}\] is the same as last two digits of \[{{4}^{8}}\]. So, the last two digits of \[{{4}^{7\times {{14}^{13}}}}\] is 36.

Hence, the last two digits of \[{{14}^{{{14}^{14}}}}\] is 36.

Note: If we will try to find the exact value of the given number, it will take a lot of time. Hence, it’s better to use tricks involving binomial expansion to find the last two digits of the given number.