The units digit of 35^87 + 93^46 is:

Easier approach would be:

The units digit of (35)^(87) is the same as the units digit of 5^(87). 5 in ANY positive integer power has the units digit of 5.

The units digit of (93)^(46) is the same as the units digit of 3^(46)=9^23. 9 in odd power has the units digit of 9.

5 + 9 = 14.

Answer: B.
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Step 1: (35)^(87) can be broken up into (5)^87 x (7)^87. Each power of 7 ends in a units digit of either a 7,9,3 or 1. Each power of 5 ends in a 5. When you multiply 5 by any odd number you will end up with a units digit of 5.

Step 2: (93)^(46) can be broken up into (3)^46 x (31)^46. Each power of 31 ends in a units digit of 1. Each power of 3 ends in a units digit of either 3,9,7 and 1. Since there is a pattern here where every 4th power of 3 ends in a units digit of 3, the 46th power of 3 would end in a units digit of 9. When you multiply 1 by 9 you end up with a units digit of 9.

Therefore, the units digit we are looking for is 5 + 9 = 14. Units digit will be 4. Answer B.

I think the answer is B. This is how I approached it..

35^87 + 93^46

Looking at 35^87 and its unit digit (5)- you know this number will end with a units digit 5 because that's how the cycle works with 5's (5^1=5, 5^2=25, 5^3-125 etc..)
Looking at 93^46 - you can establish a pattern with the 3's as I've demonstrated below...

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243 (this is where the cycle starts to repeat)

So the cycle is in 4. 46/4 leaves a remainder of 2 so you know this unit digit will end in 9.

5 + 9= 14, thus units digit is 4 and answer is (B)
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It is quite intuitive to go for a basic two step approach for this problem.

When dealing with 35^87, we can apply a simple concept here. 5 raised to any power > 0 must have 5 as its units digit.
When dealing with 93^46, we can apply the concept of cyclicity. Since the cyclicity of 3 is 4, so units digit of 93^46 is equivalent to teh units digit of 3^2 i.e. 9.
On adding these 2 digits i.e. 9 and 5, we get 14 of which the units digit is 4.
Will be curious to know how others deal with such questions.

Since we only need to determine the units digit of (35)^(87) + (93)^(46), we can rewrite the question as 5^87 + 3^46.

Let’s first determine the units digit of 5^87. Since 5 raised to any positive integer power will always have a units digit of 5, 5^87 has a units digit of 5.

Next we can determine the units digit of 3^46. 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 a power.

3^1= 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

As we can see from the above, the pattern of the units digit of any power 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:

3^44 has a units digit of 1, 3^45 has a units digit of 3, and thus 3^46 has a units digit of 9.

Since the sum of the units digits of 5^87 and 3^46 is 5 + 9 = 14, (35)^(87) + (93)^(46) has a units digit of 4.

Answer: B
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Thanks Bunuel. "9 in odd power has units digit of 9". Missed that simple logic...

MAGOOSH OFFICIAL SOLUTION:

We have to figure out each piece separately, and then add them. 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. Done.

The second term takes a little more work. We can ignore the tens digit, and just treat this base as 3. Here is the units digit patter for the powers of 3.
3^1 has a units digit of 3
3^2 has a units digit of 9
3^3 has a units digit of 7 (e.g. 3*9 = 27)
3^4 has a units digit of 1 (e.g. 3*7 = 21)
3^5 has a units digit of 3
3^6 has a units digit of 9
3^7 has a units digit of 7
3^8 has a units digit of 1

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.

Answer = B
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­Hey how is "The units digit of (93)^(46) is the same as the units digit of 3^(46)=9^23" ?

The units digit of \((...xyz)^n\) is the same as that of \(z^n\), so only the units digit of the base matters.
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Yes I got that, but how did you get 9 to the power 23?

Posted from my mobile device

 

\(3^{46}=\)

\(=3^{2*23}=\)

\(=(3^2)^{23}=\)

\(=9^{23}\)

8. Exponents and Roots of Numbers

• Theory
  Exponents and Roots on the GMAT: Tips and hints
  Cyclicity and the remainders on the GMAT
  
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    Units digits, exponents, remainders problems
    Tough and tricky DS exponents and roots questions with detailed solutions
    Tough and tricky PS exponents and roots questions with detailed solutions
    Exponents DS Questions
    Exponents PS Questions
    Roots DS Questions
    Roots PS Questions

Check below for more:
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Ultimate GMAT Quantitative Megathread

Hope this helps.­
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­That's so smart! I've been loking at some of your answer explanations on other posts and they are so helpful. Thanks a lot!