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Bumping for review and further discussion.

For more on this kind of questions check Units digits, exponents, remainders problems collection.
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3,9,7,1 repeats after every four times

therefore 350/4 = 87 + 2 remainder

9 is the ones digit

so E is coorect




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supreetb
Find the ones digit of 73^350

A) 3
B) 5
C) 6
D) 7
E) 9

My answer was A, but it turned out incorrect. Can someone help understand why.
To find unit digit of \(73^{350}\) , we need to find unit digit of \(3^{350}\)

Power of \(3\) has cyclicity of \(4\) \((3,9,7,1)\). \(\frac{350}{4}\) gives remainder of \(2\).

Therefore new power will be \(3^2\). And Unit digit of \(3^2 = 9\)

Therefore Unit digit of \(73^{350} = 9\)

Answer (E)...
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The unit digit of 73^350 is same as that of 3^350.
Power of 3 has cyclicity of 4. That means set of 4 values repeats {3,9,7,1}.
3^1=3 (3)
3^2=9 (9)
3^3=27 (7)
3^4=81 (1)
3^5=243 (3)
and cycle continues.

Hence 3^350= 3^348 * 3^2
=3^2
=9.

Hence Option E.

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supreetb
Find the ones digit of 73^350

A) 3
B) 5
C) 6
D) 7
E) 9

Since we only care about units digits, we really are determining the units digit of 3^350.

Let’s evaluate the units digit 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 a power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The pattern of the units digits 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 the units digit. Thus:

3^348 has a units digit of 1 and 3^349 has a units digit of 3, and thus 3^350 has a units digit of 9.

Answer: E
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3,9,7,1 will keep repeating

So, 350/4 = 87 + 2 remainder

9 is the ones digit

Ans->E
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supreetb
Find the ones digit of 73^350

A) 3
B) 5
C) 6
D) 7
E) 9

For unit digit, let see only 3^350 last digit.
Unit digit 3^1 = 3
Unit digit 3^2 = 9
Unit digit 3^3 = 7
Unit digit 3^4 = 1
And then it repeats, i.e after four intervals.

350 = 87*4 + 2
For 3^350, unit digit = 9 (as 2 is left after dividing by 4)

Hence, OA is (E).
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\(73^{350} = 3^ {350}\)

=> \(3^ {4 * 87} * 3^ {2}\)

=> unit digit = \(3^{2} = 9\)

Answer E
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




We need to find the units digit of \(73^{350}\)

Units digit of \(73^{350}\) will be same as units digit of \(3^{350}\)

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 (350) by 4 and check what is the remainder
350 divided by 4 gives 2 remainder

=> Units digit of \(73^{350}\) = Units digit of \(3^2\) = 9

So, Answer will be E
Hope it helps!

Watch this video to MASTER how to find Units digit of Power of 3
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