Question

Assertion :(A): The digit in the unit place of $183!+3^{183}$ is $7$. Reason: (R): $183!$ have unit place $0$ and $3^{4k+3}$ ends with $7$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A)
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

The correct option is A Both (A) & (R) are individually true & (R) is correct explanation of (A).

$183!=1\times 2\times 3\times 4\times 5\times 6\times$...$\times 182\times 183$

$=(2\times 5)(1\times 3\times 4\times 6\times ...\times 183)$

$=10\times (1\times 3\times 4\times 6\times ...\times 183)$

$=10n$ , where $n$ is a natural number.

Hence unit place of $183!$ is $0$

Now, we know $3^0=13^1=33^2=93^3=273^4=81$

So, we can observe that the unit place of $3^{4k}=1$ and that of $3^{4k+3}=7$

As, $183$ is of the form $4k+3$, $3^{183}$ has $7$ in unit place.

Hence $A$ is correct.