The thousand's digit of $3^{100}$ is

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Solution

The correct option is C $2$
$3^{100}$
$= (9)^{50}$
$= (10 - 1)^{50}$
$= 10^{50} -^{10} C_{1} 10^{49} +^{10} C_{2} 10^{48} - . . . -^{50} C_{49} 10 +^{50} C_{50} 10^{0}$
$= 10 (m) +^{50} C_{48} 10^{2} -^{50} C_{49} 10 +^{50} C_{50} 10^{0}$
$= 10 m + \frac{50.49}{2} 10^{2} - 50 (10) + 1$
$= 10 (m) + (122500) - 500 + 1$
$= 10 (m) + 122000 + 1$
$= 10 (m) + 122001$
Hence the digit at thousand's place is 2.

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The thousand's digit of 3100 is

The correct option is C 23100=(9)50=(10−1)50=1050−10C11049+10C21048−...−50C4910+50C50100=10(m)+50C48102−50C4910+50C50100=10m+50.492102−50(10)+1=10(m)+(122500)−500+1=10(m)+122000+1=10(m)+122001Hence the digit at thousand's place is 2.

The thousand's digit of $3^{100}$ is