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

A. 0

B. 2

C. 4

D. 6

E. 8

## Units digit?

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- GMATGuruNY
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3Â¹ --> units digit of 3.oquiella wrote:If N=(33)^43 + (43)^33, what is the units digit of n?

A. 0

B. 2

C. 4

D. 6

E. 8

3Â² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)

3Â³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)

3â�´ --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)

From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.

The units digits repeat in a CYCLE OF 4.

Implication:

When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus:

33â�´â�° and 43Â³Â² each have a units digit of 1.

From here, the cycle of units digits will repeat: 3, 9, 7, 1...

Thus:

33â�´Â¹ and 43Â³Â³ each have a units digit of 3.

33â�´Â² has a units digit of 9.

33â�´Â³ has a units digit of 7.

Result:

Since n = 33â�´Â³ + 43Â³Â³, n ---> (units digit of 3) + (units digit of 7) = units digit of 0.

The correct answer is A.

Last edited by GMATGuruNY on Sun Jan 07, 2018 4:16 am, edited 1 time in total.

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Then you can practice with this question - https://www.gmatprepnow.com/module/gmat- ... video/1032

Cheers,

Brent

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

A. 0

B. 2

C. 4

D. 6

E. 8

(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 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 base of 3 gives us a units digit pattern of 3-9-7-1. 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 backward one exponent in our pattern and we get:

3^43 = units digit of 7

Similarly, for the units digit of (3)^33, the closest multiple of 4 to 33 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

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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: A

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