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

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

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

First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}:
3^1=3 (the units digit is 3)
3^2=9 (the units digit is 9)
3^3=27 (the units digit is 7)
3^4=81 (the units digit is 1)
3^5=243 (the units digit is 3 again!)
...

Thus:
The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3).
The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

Hi Archit,

Just follow the unit digits:

n = (33)^43 + (43)^33

Here the last digits would determine the power.

So,

n = (33)^43 + (43)^33 ~ 3^43 + 3 ^33

Now last digits of 3^1, 2, 3, 4, 5 = 3, 9, 7, 1, 3. The cycle repeats after every 4 rounds.

So n = 3^43 + 3 ^33 = 3^(40+3) + 3^(32+1) = {(3^(4*10)}{3^3} + {(3^(4*8)}{3^1}

Now last digits of these terms would be {1}{7} + {1}{3} = 10

Hence the last digit is 0.

Hope this helps.

Regards,

Shouvik

Both terms has 3 at the units place, so the units place cycle around 3,9,7,1
33^43 will have 7 & 43^33 will have 3

7 + 3 = 10, 0 would be in the units place of the resultant

Answer = 0 = A

I can't understand why it has to be a multiple of 4.
I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong.
Generally in such exercises it has to be a multiple of 4, or I don't get something?

Based on the Number Theory > Last digit of a power (in the GMATclub forum, I cannot link to the post)

Step 1: Find the last digit of 33^43

find the cyclicity of 33
33^1= _3
33^2= _9
33^3= _7
33^4= _1
33^5= _3
-> 33 has the cyclicity of 4

Divide the power by the cyclicity: 43/4 -> remainder is 3 (refer to the 3rd position in the cyclicity)
-> the last digit of 33^43 is the same as that of 33^3 -> 7 is the last digit

Step 2: Similarly, find the last digit of 43^33
43 also has the cyclicity of 4 (ends with 3)
Divide 33 by 4 -> remainder is 1 -> refer to the 1st position in the cyclicity -> last digit is 3

Step 3: 7+3 = 10 -> the last digit is 0

Answer A

I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??

You take the remainder obtained by dividing the power by cyclicity. For example, 3^43 --> 43 (power) divided by the cyclicity of 4 gives the remainder of 3 --> the units digit of 3^43 is the same as the units digit of 3^3.

Check Units digits, exponents, remainders problems directory in our Special Questions Directory.

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 get when a base of 3 is raised to consecutive 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 base of 3 gives 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 up 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 is A.

Units digit of a product will ALWAYS depend on the units digits of the numbers being multiplied.

Just focus on the units digit.
Units digit of n=(3)^43+(3)^33
The cyclicity of 3 is 4.
3^1=3
3^2=9
3^3=27 (Units digit 7)
3^4=81 (Units digit 1)

After this the digits will start repeating (3,9,7,1,3,9,7,1, and so on)
A the cyclicity is 4 (here), write the powers as multiple of 4 and look for the remainder (if any).

43=4*10+3
Therefore the units digit of (3)^43= 7

33=4*8+1

Therefore the units digit of (3)^33= 3

The units digit of n= 7+3=10 (Just consider the last digit i.e. the unit's digit)
The answer is 0.

Look for a pattern

33^1 = 3
33^2 = (33)(33) = ---9 [aside: we need not determine the other digits. All we care about is the units digit]
33^3 = (33)(33^2) = (33)(---9) = ----7
33^4 = (33)(33^3) = (33)(---7) = ----1
33^5 = (33)(33^4) = (33)(---1) = ----3

NOTICE that we're back to where we started.
33^5 has units digit 3, and 33^1 has units digit 3
So, at this point, our pattern of units digits keep repeating 3, 9, 7, 1, 3, 9, 7, 1, 3,...
We say that we have a "cycle" of 4, which means the digits repeat every 4 powers.

Now that we know the pattern has a cycle of 4, let's examine powers where the exponent is a multiple of 4
We know that 33^4 has units digit 1
So this means 33^8 has units digit 1
And 33^12 has units digit 1
And 33^16 has units digit 1
And so on

To find the unit's digit of (33)^43, let's find the nearest multiple of 4 that's less than 43.
40 is a multiple of 4, which means 33^40 has units digit 1
From here we can continue our pattern to see that
33^41 has units digit 3
33^42 has units digit 9
33^43 has units digit 7
--------------------------------------

Since 33 and 43 have the same units digit, 33 and 43 will have the exact same cycle of 4
So, to find the unit's digit of (43)^33, we'll first find the nearest multiple of 4 that's less than 33.
32 is a multiple of 4, which means 43^32 has units digit 1
From here we can continue our pattern to see that
43^33 has units digit 3
---------------------------------------

So, (33)^43 + (43)^33 = ----7 + ----3 = ----0

Answer: A

Here's an article I wrote on this topic (with additional practice questions): https://www.gmatprepnow.com/articles/un ... big-powers

Cheers,
Brent

Solution

The unit digit depends on the unit digit of 3^43 and 3^33

Unit digit of 3^43 -

Cyclicity of 3 is 4

=>43 divided by 4 gives the remainder of 3

=> Unit digit of 3^43 = Unit digit of 3^3

=7

Unit digit of 3^33 -

=>33 divided by 4 gives the remainder of 1

=> Unit digit of 3^33 = Unit digit of 3^1

=3

Thus the units digit of (33)^43 + (43)^33

= 7 + 3 = 0 (option a)

Devmitra Sen(GMAT Quant Expert)

Hi All,

We’re told that N = 33^43 + 43^33. We’re asked for the UNITS DIGIT of N. The GMAT would NEVER expect you to calculate that type of incredibly large result, so there MUST be a pattern involved in those numbers. When dealing with large Exponents, it is likely that there will be a repeating pattern in the Units digit, so we should do enough work to define what that pattern is… and then we’ll use that pattern to answer this question.

The phrase “units digit” is a technical term that means “ones digit.” For example, in the number 127, the number ‘7’ is the units digit.

When multiplying a number by itself again-and-again, if you are only interested in the units digit, then you don’t really have to pay attention to anything else. For example:

(33)(33)… when multiplying here, the units digit will be the PRODUCT of the two units digits, meaning (3)(3) = 9 will be the units digit.

(33)(33)(33)… we already know that 33^2 has a units digit of 9, so I’m going to refer to everything BUT the 9 with place-holders…
(~~~9)(33) … (9)(3) = 27, so the units digit of (33)^3 is a 7.

Now that we’ve defined how the math ‘works’, we can figure out the pattern when multiplying all of those 33s together and all of those 43s together.

(33)^1 … ends in a 3
(33)^2 … ends in a 9
(33)^3 … ends in a 7
(33)^4 … ends in a 1
(33)^5 … ends in a 3

Notice how the cycle is 3…9…7…1 and then it repeats with a 3… (which will then be followed by 9..7..1 over-and-over). THIS is the pattern that we’re looking for.

33^43 is ten ‘sets’ of ‘3971’ with three more 33s multiplied in. The units digit here is 7.

43^33 is eight ‘sets’ of ‘3971’ with one more 43 multiplied in. The units digit here is 3.

When you add a number that ends in a 7 with a number that ends in a 3, you get a number that ends in a 0.

Final Answer:

GMAT Assassins aren’t born, they’re made,
Rich

Video solution from Quant Reasoning:
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­In order to be sure for the understanding of the method, if we had to find the unit digit of 47^26 + 26^47 we would focus on the 7^6 which gives us unit digit 9 and 6^7 with unit digit 9. So the unit digit of 47^26 + 26^47 is 9+6 = 15, am I correct?
 

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