Question

Find the last digit, last two digits and last three digits of the number \[{{(27)}^{27}}\].

Answer 1

Hint: We will first convert \[{{(27)}^{27}}\] to a binomial form and then we will use the binomial expansion formula \[{{(a+b)}^{n}}={{a}^{n}}+({}^{n}{{C}_{1}}){{a}^{n-1}}b+({}^{n}{{C}_{2}}){{a}^{n-2}}{{b}^{2}}+........+({}^{n}{{C}_{n-1}})a{{b}^{n-1}}+{{b}^{n}}\] and after this we will divide the whole expression by 100 for finding last two digits and by 1000 for finding the last three digits.

Complete step-by-step answer:
We can write \[{{(27)}^{27}}={{({{3}^{3}})}^{27}}={{3}^{81}}\]. And we know that cyclicity of 3 is 4. Hence 81 divided by 4 leaves a remainder of 1. Therefore, the last digit will be 3.
We can write 3 to the power 81 also as,
\[\Rightarrow {{3}^{81}}=3\times {{3}^{80}}=3\times {{({{3}^{2}})}^{40}}=3\times {{9}^{40}}.........(1)\]
So now we can write 9 as 10 minus 1 in equation (1) so we get,
\[\Rightarrow 3\times {{(10-1)}^{40}}.........(2)\]
Now we know the binomial expansion formula is,\[{{(a+b)}^{n}}={{a}^{n}}+({}^{n}{{C}_{1}}){{a}^{n-1}}b+({}^{n}{{C}_{2}}){{a}^{n-2}}{{b}^{2}}+........+({}^{n}{{C}_{n-1}})a{{b}^{n-1}}+{{b}^{n}}\]
So applying this formula in equation (2) we get,
\[\Rightarrow 3\times ({{10}^{40}}-{}^{40}{{C}_{1}}\times {{10}^{39}}+........+{}^{40}{{C}_{38}}\times {{10}^{2}}-{}^{40}{{C}_{39}}\times 10).........(3)\]
Now for finding the last two digits we will divide the whole expression in equation (3) by 100. And we can see each term in the above expression in equation (3) contains 100 except 1 terms and assuming t to be an integer. So doing all this and using the combination formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}\], we get,
\[\Rightarrow {{3}^{81}}=3\times (100t+1).....(4)\]
Now solving equation (4) we get,
\[=300t+3.....(5)\]
So from equation (5) we can see that as the last two digits of 300t will be 00. So now adding 3 the number will be ……….00+3 equal to …….03. Hence the last two digits is 03.
Now for finding the last three digits we will divide the whole expression in equation (3) by 1000. And we can see each term in the above expression in equation (3) contains 1000 except \[{}^{40}{{C}_{39}}\times 10\] and 1. Assuming x to be an integer. So doing all this and using the combination formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}\], we get,
\[\Rightarrow {{3}^{81}}=3\times (1000x-{}^{40}{{C}_{39}}\times 10+1).....(6)\]
Now solving equation (6) we get,
\[\begin{align}
  & =3\times (1000x-\dfrac{40!}{1!\times 39!}+1) \\
 & =3\times (1000x-400+1) \\
 & =3\times (1000x-399) \\
 & =3000x-1197.........(7) \\
\end{align}\]
So from equation (7) we can see that as the last two digits of 3000x will be 000. So now subtracting 1197, the number will be ……….000-1197 equal to …….803. Hence the last three digits is 803.

Note: Cyclicity is the repetitive pattern of being divided by a certain number. We know \[{{3}^{1}}=3\], \[{{3}^{2}}=9\], \[{{3}^{3}}=27\] and \[{{3}^{4}}=81\] and after these it starts repeating. Hence the cyclicity of 3 is 4. Also remembering the binomial expansion formula and combination formula is the key here. We in a hurry can commit a mistake in applying the expansion formula in equation (2) so we need to be careful while doing this step.