Question

What is the unit digit number of 2^394x3^2459​

Answer 1

Answer:

$\huge \colorbox{lightblue} {Answer}$

$= {2}^{394} \times {3}^{2459}$

First we will find the last digit of

$First \: \: we \: will \: find \: \: the \: last \: digit \: of \: {2}^{394}$

Cyclicity of 2 is 4..

We will divide 394 by 4 we will get 2 as remainder so it means that

$= {2}^{394} = {2}^{4n + 2} \\ then \: last \: digit \: is \: 4.$

$now\: \: we \: will \: find \: \: the \: last \: digit \: of \: {3}^{2459} \\ = cyclicity \: of \: 3 \: is \: 4.. \\ = We \: will \: divide \: \: 2459 \: by \: 4 \: we \: will \: get \: 2 \: as \: remainder \: so \: it \: means \: that \: \\ = {3}^{2459} = {3}^{4n + 2} \\ then \: last \: digit \: is \: 9.$

$= so \: now \: the \: last \: digit \: of \: {2}^{394} is \: 4 \\ = the \: last \: digit \: of \: {3}^{2459} is \: 9.$

Product of 4 and 9 is 36..

$= unit \: digit \: of \: {2}^{394} \times {3}^{2459} is \: 6$

Answer 2

Answer:

Unit Digit is 6

Step-by-step explanation:

Hope this helps