Question

Pls find the tens digit of 3^2016

Answer 1

$It \: is \: very \: tough \: and \: time \: consuming \: to \: find \: the \: value \: of \: {3}^{2016}$

$we \: know \: that:$

●The ones place of any number raised to any number enters a loop like value.

For instance let's take 3:

${3}^{0} = 1. \\ {3}^{1} = 3. \\ {3}^{2} = 9. \\ {3}^{3} = 27. \\ \\ and \: now \: concentrate \: on \: the \: ones \: place: \\ {3}^{4} = 81. \\ here \: we \: can \: see \: that \: the \: ones \: place \: digit \: has \: started \: repeating \: and \: so \: the \: loop \: is \: like:$
1,3,9,7 and so it repeats itself.

Now we can't use this as we need to find the tens place.

So, here I'm writing the values till 3^n where 'n' is the exponent when it enters into the loop.

${3}^{0} = 01 \\ {3}^{1} = 03 \\ {3}^{2} = 09 \\ {3}^{3} = 27. \\ {3}^{4} = 81 \\ \\ after \: this \: the \: number \: becomes \: a \: 3 - digit \: number \: and \: i \: am \: going \: to \: write \: only \: the \: ones \: and \: tens \: place \: of \: them \\ \\ {3}^{5} = 43 \\ {3}^{6} = 29 \\ {3}^{7} = 87 \\ {3}^{8} = 61 \\ {3}^{9} = 83 \\ {3}^{10} = 49 \\ {3}^{11} = 47 \\ {3}^{12} = 41 \\ {3}^{13} = 23 \\ {3}^{14} = 69 \\ {3}^{15} = 07 \\ {3}^{16} = 21 \\ {3}^{17} = 63 \\ {3}^{18} = 89 \\ {3}^{19} = 67 \\ {3}^{20} = 01$
Now that 01 has repeated, we can say that the tens place loops after every 20 powers.

So, we can find the tens place of 3^2016 by dividing 2016 by 20 and finding the remainder obtained.

$\frac{2016}{20} = 100 \frac{16}{20}$

Here, the remainder is 16.

So, we can say that the ten's place of 3^2016 is the same as 3^16's ten's place value.

Therefore, 2 is the value of the ten's place of 3^2016.