Problem Solving

If n = (33)^43 + (43)^33, what is the units digit of n ?

(A) 0
(B) 2
(0 4
(D) 6
(E) 8

These kinds of questions are discussed in our free video: https://www.gmatprepnow.com/module/gmat- ... ts?id=1031

To begin, look for a pattern.
33^1 = 33 (units digit is 3)
33^2 = --9 (units digit is 9)
Aside: we need only determine the units digit, so don't worry about the other digits
33^3 = --7 (units digit is 7)
33^4 = --1 (units digit is 1)
33^5 = --3 (units digit is 3)
33^6 = --9 (units digit is 9)
33^7 = --7 (units digit is 7)
33^8 = --1 (units digit is 1)
etc.
As you can see, the pattern repeats every 4 exponents.
So, we say the pattern has a cycle of 4.

Also notice that 33^4 has units digit 1, 33^8 has units digit 1, 33^12 has units digit 1, etc.
So, 33^40 will have units digit 1. Continuing with the pattern...
33^41 has units digit 3
33^42 has units digit 9
33^43 has units digit 7

We can apply the same technique to see that 43^33 has units digit 3

So, (33)^43 + (43)^33 will have units digit 0, since 7 + 3 = 10 [spoiler](units digit 0)[/spoiler]

Answer: A

You can also read my BTG article on this topic: https://www.beatthegmat.com/mba/2012/10/ ... big-powers

Cheers,
Brent

hayes_meredith88 wrote:I keep getting caught up on this properties question in the 13th Edition Official Guide. #163 in the problem solving practice:

if n= 33^43 + 43^33, what is the units digit of n?

A)0
B)2
C)4
D)6
E)8

The correct answer is A)0. I understand building the sequence...3,9,7,1,3,9,7 etc. I get stuck understanding the reasoning behind the explanation "since 43= (10(4)+3, the 43rd number in this sequence is 7". How is that determined? Please help!

Solution:

This is a units digit pattern question. The first thing to recognize is that in units digit pattern questions we only care about the units digit place value. Thus, we can rewrite the problem as:

(3)^43 + (3)^33

We now need to determine the units digit of (3)^43 + (3)^33. Let's determine the pattern of units digits that we can get when a base of 3 is raised to positive integer exponents.

3^1 = 3

3^2 = 9

3^3 = 27 (units digit of 7)

3^4 = 81 (units digit of 1)

3^5 = 243 (units digit of 3)

Notice at 3^5, the pattern has started over:

3^6 = units digit of 9

3^7 = units digit of 7

3^8 = units digit of 1

So we can safely say that the powers of 3 give us a units digit pattern of 3, 9, 7, 1, 3, 9, 7, 1,... that repeats every four exponents. Also notice that every time 3 is raised to an exponent that is a multiple of 4, we are left with a units digit of 1. This is very powerful information, which we can use to solve the problem. Let's start with the units digit of (3)^43.

An easy way to determine the units digit of (3)^43, is to find the closest multiple of 4 to 43, and that is 44. Thus we know:

3^44 = units digit of 1

So we can move back one exponent in our pattern and we get:

3^43 = units digit of 7

Let's now determine the units digit of (3)^33.

We already know that the pattern of units digits for powers of 3 will be 3, 9, 7, 1, 3, 9, 7, 1,... An easy way to determine the units digit of (3)^33 is to find the closest multiple of 4 to 33, and that is 32. Thus we know:

3^32 = units digit of 1

So we can move forward one exponent in our pattern and we get:

3^33 = units digit of 3

The last step is to add the two units digits together so we have:

7 + 3 = 10, which has a units digit of zero

Answer: A