Question

Find the last two digits of ${41}^{2786}$

Answer 1

To calculate last two digits of a number, we use binomial theorem:

$(x+a{)}^n={}^n{\text{c}}_0{a}^n+{}^n{\text{c}}_1{a}^{n-1}x+{}^n{\text{c}}_2{a}^{n-2}{x}^2+.........$

where ${}^n{\text{c}}_r=\frac{n!}{r!(n-r)!}$

According to question,

$(41{)}^{2786}=(40+1{)}^{2786}$

$={}^{2786}{\text{c}}_0{1}^{2786}+{}^{2786}{\text{c}}_1{1}^{2785}\times 40+{}^{2786}{\text{c}}_2\times 1^{2784}(40{)}^2+.........$

Note that all the terms after the second term will end in two or more zeroes.

The first two terms are ${}^{2786}{\text{c}}_0\times 1^{2786}\text{and}{}^{2786}{\text{c}}_1\times 1^{2785}\times 4$

Now, the second term will end with one zero.

tens digit of the second term will be product of 2786 & 40 i.e. 6. The last digit of first term is 1.

So the last two digits of ${41}^{2786}$ are 61.