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check for power 333
3 has a cyclic repetition of 4. so unit digit of \(3^1\) is same as\(3^5\)
so lets divide 333 by 4.
this leaves a remainder of 1. Hence \((333)^(333)\) will be same as \(3^1\) = 3.
Thus B is the answer
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Bunuel
Find the units digit of 333^333

A. 1
B. 3
C. 5
D. 7
E. 9

For \(3^n\), unit digit will have a cyclicity pattern as 3 9 7 1

4|333|83, remainder is 1, signifies 1st position

Answer B
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Bunuel
Find the units digit of 333^333

A. 1
B. 3
C. 5
D. 7
E. 9

Since we care only about units digits, we can rewrite the expression as:

3^333

We can evaluate the pattern of the units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to each power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The pattern of the units digit of powers of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

3^332 has a units digit of 1, so 3^333 has a units digit of 3.

Answer: B
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333^333
=> 3^333 * 111^333

cyclic nature of 3^333 would give us units as 3( 3, 9, 7, 1)
and 1^333; 1
so 3*1 = 3

Ans->B
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



We need to find the units digit of \(333^{333}\)

Units digit of \(333^{333}\) will be same as units digit of \(3^{333}\)

Lets start by finding the cyclicity of units' digit in powers of 3

\(3^1\) units’ digit is 3
\(3^2\) units’ digit is 9
\(3^3\) units’ digit is 7
\(3^4\) units’ digit is 1
\(3^5\) units’ digit is 3

That means that units digit of power of 3 has a cycle of 4

=> We need to divide the power (333) by 4 and check what is the remainder
333 divided by 4 gives 1 remainder

=> Units digit of \(333^{333}\) = Units digit of \(3^1\) = 3

So, Answer will be A
Hope it helps!

Watch these videos to MASTER how to find Units digit of Exponents
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