Question

Find (i) the last digit, (ii) the last two digits, and (iii) the last three digits of\[{17^{256}}\].

Answer 1

Hint- Use the binomial theorem expansion to find the number having higher powers. Expand the number in the form of \[\left( {a + b} \right)\] to find the number of digits. Binomial theorem expansion is an easy method to find the expansion having a higher rise to a power. To find the numbers having raised to the \[2,3,4,5.....10\]power could be an easy task, but having raised to the power of\[200,300,400,500,......1000\]becomes lengthy and time-consuming; hence, binomial theorem expansion makes it easy.
Binomial theorem expression is written as:
\[{\left( {a - b} \right)^n} = {\left( { - 1} \right)^n}\sum\limits_{k = 0}^n {\left( {{}^n{C_k}} \right){a^{n -
k}}{b^k}} \]Where, \[{}^n{C_k} = \dfrac{{n!}}{{\left( {n - k} \right)!k!}}\]
Here, \[n!\] is a factorial notation which is the product of value between \[1\] and \[n\],


Complete step by step solution:
We can write \[{17^{256}}\]as \[{17^{256}} = {17^{\left( {2 \times 128} \right)}} = {\left( {{{17}^2}}
\right)^{128}} = {\left( {289} \right)^{128}} = {\left( {290 - 1} \right)^{128}}\]
Now use binomial theorem expansion to expand the given number as:
\[{\left( {290 - 1} \right)^{128}} = \left[ {{}^{128}{C_0}{{\left( {290} \right)}^{128}} - {}^{128}{C_1}{{\left(
{290} \right)}^{127}} + {}^{128}{C_2}{{\left( {290} \right)}^{126}} - {}^{128}{C_3}{{(290)}^{125}} + ..............
- {}^{128}{C_{125}}{{\left( {290} \right)}^3}} \right] + {}^{128}{C_{126}}{\left( {290} \right)^2} -
{}^{128}{C_{127}}{\left( {290} \right)^1} + {}^{128}{C_{128}}{\left( {290} \right)^0}\]
In the above expansion, we will divide the expansion into two different groups
\[{\left( {290 - 1} \right)^{128}} = \left[ {{}^{128}{C_0}{{\left( {290} \right)}^{128}} - {}^{128}{C_1}{{\left(
{290} \right)}^{127}} + {}^{128}{C_2}{{\left( {290} \right)}^{126}} - {}^{128}{C_3}{{(290)}^{125}} + ..............
- {}^{128}{C_{125}}{{\left( {290} \right)}^3}} \right] + {}^{128}{C_{126}}{\left( {290} \right)^2} -
{}^{128}{C_{127}}{\left( {290} \right)^1} + {}^{128}{C_{128}}{\left( {290} \right)^0}\]
As \[{}^n{C_k} = \dfrac{{n!}}{{\left( {n - k} \right)!k!}}\] we can write.
\[{}^{128}{C_{126}} = \dfrac{{128!}}{{\left( {128 - 126} \right)!126!}} = \dfrac{{128!}}{{126! \times 2!}} =
\dfrac{{128 \times 127}}{2}\]
\[{}^{128}{C_{127}} = \dfrac{{128!}}{{\left( {128 - 127} \right)!127!}} = \dfrac{{128!}}{{127! \times 1!}} =
128\]
\[{}^{128}{C_{128}} = \dfrac{{128!}}{{\left( {128 - 128} \right)!128!}} = \dfrac{{128!}}{{128! \times 0!}} =
1\]

Also, \[0! = 1\]. Since factorial is the product of the whole number between \[1\] to \[n\]hence factorial
of \[0\] starts with \[1\]so the value is \[1\].
Now
\[{\left( {290 - 1} \right)^{128}} = \left[ {{}^{128}{C_0}{{\left( {290} \right)}^{128}} - {}^{128}{C_1}{{\left(
{290} \right)}^{127}} + {}^{128}{C_2}{{\left( {290} \right)}^{126}} - {}^{128}{C_3}{{(290)}^{125}} + ..............
- {}^{128}{C_{125}}{{\left( {290} \right)}^3}} \right] + \dfrac{{128 \times 127}}{2}\left( {290 \times 290}
\right) - 128 \times 290 + 1\]The terms under the bracket that we marked we can write \[{290^3}\]as:
\[{290^3} = {29^3} \times {10^3} = {29^3} \times 1000\]
In the above group, each element is to the multiple of 1000; hence we can write the group as 1000k
where k is an integer. So,
\[
{\left( {290 - 1} \right)^{128}} = 1000k + \dfrac{{128 \times 127}}{2}\left( {290 \times 290} \right) - 128
\times 290 + 1 \\
= 1000k + 683564800 - 36250 + 1 \\
= 1000k + 683527681 \\
\]
We can also write
\[
1000k + 683527681 = 1000k + 683527000 + 681 \\
= 1000\left( {m + 683527} \right) + 681 \\
\]
Hence we have \[681\]as the last three digits of \[{17^{256}}\]
Last digit\[ = 1\]
Last two digits\[ = 81\]
Last three digits\[ = 681\]

Note: Generally, when we are asked to find the digits of the numbers having a higher power of them, we use Binomial theorem expansion. Alternatively, the problem can be solved by dividing the raised power by 4 and calculating the remainder. If the remainder is 1 then, the last digit is the last digit of the number itself; if the remainder is 2, then the last digit will be the last digit of the square of the number, while if the remainder is 3, then the last digit will be the last digit of the cube of the number.