Quiz Level-3

Hint

Four to any odd power will have a units digit of 4 . Thus, any number with a units digit of four, raised to an odd power, will also have a units digit of 4 . The first factor, \(44^{91}\), has a units digit of 4.
Thus, any number with a units digit of 3, when raised to the power of 37 , will have a units digit of 3. The second factor, \(73^{37}\), has a units digit of 3.
Of course, 4 x 3 = 12, so any number with a units digit of 4 times any number with a units digit of 3 will yield a product with a units digit of 2.

Hint

The first piece is remarkably easy — any power of anything ending in 5 always has a units digit of 5.  So the first term has a units digit of 5. 

The period is 4 . This means, 3 to the power of any multiple of 4 will have a units digit of 1. \(3^{44}\) has a units digit of 1 \(3^{45}\) has a units digit of 3 \(3^{46}\) has a units digit of 9.

Therefore, the second term has a units digit of 9.

Of course 5 + 9 = 14, so something with a units digit of 5 plus something with a units digit of 9 will have a units digit of 4.

Hint

\(
1 !=1
\)
\(
2 !=2
\)
\(
3 !=6
\)
\(
4 !=24
\)
\(
5 !=120
\)
\(
6 !=620 \ldots . .
\)
The unit digit of factorial of any number greater than 4 is 0. Hence,
Unit digit of \(1 !+2 !+3 !+\ldots \ldots+88 !+89 !=\) Unit digit of \(1+2+6+4+0+0+\ldots \ldots+0\)
\(=\) Unit digit of \(13=3\)

Hint

Power series of 7 i.e units digit 7 power expansion has \(7,9,3\) and 1 and it is raised to power 413 i.e \(413 / 4\) remainder 1 and so last digit is 7
Power series of 9 i.e units digit 9 power expansion has 9 and 1 and it is raised to power 547 i.e \(547 / 2\) remainder 1 and so last digit is 9
Power series of 4 i.e units digit 4 power expansion has 4 and 6 and it is raised to power 624 i.e \(624 / 2\) remainder 0 and so last digit is 6
Power series of 2 i.e units digit 2 power expansion has \(2,4,8\) and 6 and it is raised to power 812 i.e \(812 / 4\) remainder 0 and so the last digit is 6
All the last digits product \(=7 \times 9 \times 6 \times 6\)
\(=8\)

Hint

If we multiply 433 and 456 then we will get 8 as unit digit.
But when 433 and 456 multiply together with \(43 \mathrm{~N}\) then the unit digit appears as \(8 \mathrm{~N}\).
So,Unit digit of \(8 N=N+2\).
It is possible only when \(\mathrm{N}=6\).

Hint

The units digits in the powers of 2 repeat: 2-4-8-6-2-4-8-6-…. and 222/4 leaves a remainder of 2. This means that \(222^{222}\) must end in a 4. Multiplying this result by 9 would give us a unit digit of 6.

Hint

Clearly, unit’s digit in the given product \(=\) unit’s digit in \(7^{153} \times 1^{72}\). Now, \(7^4\) gives unit digit 1.
\(\therefore 7^{152}\) gives unit digit 1 .
\(\therefore 7^{153}\) gives unit digit \((1 \times 7)=7\). Also, \(1^{72}\) gives unit digit 1 .
Hence, unit digit in the product \(=(7 \times 1)=7\).

Hint

Required unit’s digit \(=\) unit’s digit in \((4)^{102}+(4)^{103}\).
Now, \(4^2\) gives unit digit 6.
\(\therefore(4)^{102}\) gives unit digit 6.
\((4)^{103}\) gives unit digit of the product \((6 \times 4)\) i.e., 4 .
Hence, unit’s digit in \((264)^{102}+(264)^{103}=\) unit’s digit in \((6+4)=0\).

Hint

Let \(x=2 y\). Then, \(x^{4 n}=(2 y)^{4 n}=\left\{(2 y)^4\right\}^n=\left(16 y^4\right)^n\). \(y=1,2,3,4,5,6,7,8,9\) gives unit digit as 6 in \(\left(16 y^4\right)^n\).
But, \(y=5\), gives unit digit 0 in \(\left(16 y^4\right)^n\).
Hence, the unit digit is 0 or 6 .

Hint

In \(5^n\) we have 5 as unit digit and in \(6^m\) we have 6 as unit digit.
\(\therefore\) Unit digit in \(\left(5^n+6^m\right)=\) Unit digit in \((5+6)=\) Unit digit in \(11=1\).