Quiz Level-2

Hint

An integer that is divisible by 2 but not 5 must end in a 2,4,6, or 8. \((n^2)^2=n^4\), and \(2^4,4^4,6^4, 8^4\) all end in a 6, so the units digit of any integer \((n^2)^2\) where \(n\) is divisible by 2 but not 5 will always end in a 6.

Hint

These repeat every-other digit. Odd powers of a number ending in 4 end in a 4 , while even powers end in a 6.

In this case unit digit is 4 with an odd power(35), so answer is 4.

Hint

Odd powers of a number ending in 9 end in a 9 , while even powers end in a 1. In this case unit digit, 9 has an even power(52), so the answer is 1.

Hint

You can solve this by finding the remainder when 777 is divided by 4. The remainder is 1, so the units digit will be the first in the cycle which is 8.

For a number ending in \(2,3,7\), or 8 , the units digits cycle in sets of four.

\(2^1=2,2^2=4,2^3=8,2^4=16\), multiplying by 2 again we get \(2^5=32\), and the cycle of units digits repeats. The cycle for 2 ‘s is \(2-4-8-6-2-4-8-6\)…

\(3^1=3,3^2=9,3^3=27,3^4=81\), multiplying by 3 again we get \(3^5=243\), and the cycle of units digits repeats. The cycle for 3 ‘s is \(3-9-7-1-3-9-7-1 \ldots\)

The cycle for 7’s is \(7-9-3-1-7-9-3-1 \ldots\)

The cycle for 8 ‘s is \(8-4-2-6-8-4-2-6 \ldots\)

Hint

The units digit of \(1976^{61}\) is 6 (any number ending in 6 raised to a power ends in 6). The units digit of \(2007^{61}\) is 7 (the digits cycle in sets of four 7-9-3-1, and 60 is divisible by 4 , so \(2007^{60}\) ends in a 1 and \(2007^{61}\) ends in a 7\()\). The units digit of \(\left(1976^{61}\right)\left(2007^{61}\right)\) will be the same as the units digit of \(6 \cdot 7: 2\)

Hint

The units digits of the powers of 3 cycle in sets of 4 : \(3-9-7-1-3-9-7-1 \ldots\) so to find the units digit for the sum of four consecutive powers of 3, we add \(3+9+7+1=20\). The units digit is 0.

Hint

Units digits of the powers of 9 cycle \(1-9-1-9 \ldots\), so consecutive powers of 9 will end in a 1 and a 9, meaning the sum of consecutive powers of 9 will always end in a zero: \(\ldots 1+\ldots 9=\ldots 0\).

Hint

The given product is \(\left(7493^{263}\right) \times\left(151^{29}\right)\)
Required unit digit = the unit digit of \(\left(3^{263}\right) \times\left(1^{29}\right) \ldots(1)\)
In the value of 3 to the power 4 , we have the unit digit as 1 .
so, we can rewrite \(\left(3^{263}\right)=\left[3^{(4 \times 65+3)}\right]=\left[\left(3^4\right)^{65}\right] \times\left(3^3\right)\)
Then from eqn(1),
The unit digit of \(\left(3^{263}\right) \times\left(1^{29}\right)=\) The unit digit of \(\left.\left[3^4\right)^{65}\right] \times\left(3^3\right) \times 1^{29}\)
\(=\) The unit digit of \(\left[1^{65}\right] \times 27 \times 1\)
\(=\) The unit digit of \(1 \times 7 \times 1=7\)
Hence, the answer is 7.

Hint

Given that \(634^{262}+634^{263}\)
\(
\begin{aligned}
& =634^{262}(1+634) \\
& =\left(634^{262}\right) \times 635
\end{aligned}
\)
The unit digit of \(\left(634^{262}\right) \times 635=\) the unit digit of \(\left(4^{262} \times 5\right)\)
We know that the unit digit of 4 to the power of any odd number is 4 and the unit digit of 4 to the power of any even number is 6.
Then the unit digit of \(\left(4^{262} \times 5\right)=\) unit digit of \((6 \times 5)=0\)

Hint

\(
\text { 1) First of all, all we need is the last digit of the base, not } 137 \text {, but just } 7 .
\)
The period is 4 , so 7 to the power of any multiple of 4 has a units digit of 1
\(7^{12}\) has a units digit of 1
\(7^{13}\) has a units digit of 7
So the inner parenthesis is a number with a units digit of 7.
\(7^{47}\) has a units digit of 3
So the unit digit of the final output is 3 .