What is the decimal place digit of \(5^{10}\) ?
2
The decimal place digit here means the digit in the tens place, that is, the second digit from the right. To find it we only need the last two digits of the power. Powers of 5 follow a fixed pattern: from \(5^{2}\) onwards every power of 5 ends in 25, because multiplying a number ending in 25 by 5 again gives a number ending in 25.
Working:
\(5^{1} = 5\)
\(5^{2} = 25\)
\(5^{3} = 125\)
\(5^{4} = 625\)
\(5^{5} = 3125\)
\(5^{10} = (5^{5})^{2} = 3125 \times 3125 = 9765625\)
Last two digits of \(9765625 = 25\)
Tens place digit = 2
The digit 5 is the units digit of 9765625, not the tens digit, so it does not fit. The digits 7 and 0 do not appear in the tens place of any power of 5 from \(5^{2}\) onwards, since all such powers end in 25. Only the digit 2 matches the tens place of \(5^{10}\).
Hence, 2.
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