The unit's digit of the number 6^{256} - 4^{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the unit's digit of $6^{256} - 4^{256}$,we analyze the cyclicity of the powers of $6$ and $4$.For any positive integer $n$,the unit's digi

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) To find the unit's digit of $6^{256} - 4^{256}$,we analyze the cyclicity of the powers of $6$ and $4$.For any positive integer $n$,the unit's digit of $6^n$ is always $6$ because $6^1 = 6$,$6^2 = 36$,$6^3 = 216$,etc.For the powers of $4$,the unit's digit follows a cycle of length $2$: $4^1 = 4$,$4^2 = 16$ (unit digit $6$),$4^3 = 64$ (unit digit $4$),and so on.If the exponent is odd,the unit's digit is $4$. If the exponent is even,the unit's digit is $6$.Since $256$ is an even number,the unit's digit of $4^{256}$ is $6$.Therefore,the unit's digit of $6^{256} - 4^{256}$ is the unit's digit of $(6 - 6) = 0$.

The unit's digit of the number $6^{256} - 4^{256}$ is

Similar Questions

What approximate value will come in the place of the question mark $(?)$ in the following question? (You are not expected to calculate the exact value.)
$(32.13)^{2} + (23.96)^{2} - (17.11)^{2} = ?$

Which of the following is $TRUE$?
$I.$ $\sqrt[3]{11} > \sqrt{7} > \sqrt[4]{45}$
$II.$ $\sqrt{7} > \sqrt[3]{11} > \sqrt[4]{45}$
$III.$ $\sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
$IV.$ $\sqrt[4]{45} > \sqrt{7} > \sqrt[3]{11}$

$\sqrt{5^{2} \times 14 - 6 \times 7 + (4)^{?}} = 18$

What should come in place of the question mark $(?)$ in the following expression?
$820.15 + 2379.85 + 140.01 \times 4.99 = ?$

If $x = 3 - 2\sqrt{2}$,then $\sqrt{x} + \frac{1}{\sqrt{x}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module