Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: What is the units digit of the product (32^28) (33^47) (37^19)?
[#permalink]

Show Tags

10 Apr 2017, 08:55

4

Top Contributor

3

GMATPrepNow wrote:

What is the units digit of the product (32^28)(33^47)(37^19)?

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

-----ASIDE----------------------------------------- There are some "nice" numbers that, when raised to various powers, ALWAYS have the same units digit. For example, the units digit of 70^n will be 0 FOR ALL POSITIVE INTEGER VALUES OF N Likewise, the units digit of 91^n will be 1 FOR ALL POSITIVE INTEGER VALUES OF N And the units digit of 86^n will be 6 FOR ALL POSITIVE INTEGER VALUES OF N -----NOW ONTO THE QUESTION-----------------

Notice that the exponent 47 is equal to the SUM of the other two exponents (28 and 19) So, it might be useful to take 33^47 and REWRITE it as (33^28)(33^19) NOTE: later on, we'll apply a nice exponent rule that says "(a^n)(b^n) = (ab)^n"

We get: (32^28)(33^47)(37^19) = (32^28)(33^28)(33^19)(37^19) = (32^28 x 33^28)(33^19 x 37^19) = (32 x 33)^28 (33 x 37)^19 [applied above rule] =(---6)^28 (---1)^19 [I'm focusing solely on the units of each product. So, I use "---" to represent the other digits] =(----6)(----1) [When ----6 is raised to any power the units digit is always 6. The same applies to ----1] = -------6

Re: What is the units digit of the product (32^28) (33^47) (37^19)?
[#permalink]

Show Tags

09 Apr 2017, 09:24

2

32^28 for powers of 32, the units digit follows the pattern (^power - units digit): ^1 - 2 ^2 - 4 ^3 - 8 ^4 - 6 ^5 - 2 and the pattern repeats, every 4 powers we are looking for the 28th power, 28/4 = 7 R0, therefore we use the 4th place in the pattern --> 6

33^47 for powers of 33, the units digit follows the pattern (^power - units digit): ^1 - 3 ^2 - 9 ^3 - 7 ^4 - 1 ^5 - 3 and the pattern repeats, every 4 powers we are looking for the 47th power, 47/4 = 11 R3, therefore we use the 3rd place in the pattern --> 7

37^19 for powers of 37, the units digit follows the pattern (^power - units digit): ^1 - 7 ^2 - 9 ^3 - 3 ^4 - 1 ^5 - 7 and the pattern repeats, every 4 powers we are looking for the 19th power, 19/4 = 4 R3, therefore we use the 3rd place in the pattern --> 3

notice that the largest exponent is equal to the sum of the other two exponents

Cheers, Brent

Hi Brent,

How does "notice that the largest exponent is equal to the sum of the other two exponents" Helps? All(2, 3, 7, 8) the 3 numbers have cycle of 4 repetitions.

Re: What is the units digit of the product (32^28) (33^47) (37^19)?
[#permalink]

Show Tags

11 Apr 2017, 07:24

GMATPrepNow wrote:

GMATPrepNow wrote:

What is the units digit of the product (32^28)(33^47)(37^19)?

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

-----ASIDE----------------------------------------- There are some "nice" numbers that, when raised to various powers, ALWAYS have the same units digit. For example, the units digit of 70^n will be 0 FOR ALL POSITIVE INTEGER VALUES OF N Likewise, the units digit of 91^n will be 1 FOR ALL POSITIVE INTEGER VALUES OF N And the units digit of 86^n will be 6 FOR ALL POSITIVE INTEGER VALUES OF N -----NOW ONTO THE QUESTION-----------------

Notice that the exponent 47 is equal to the SUM of the other two exponents (28 and 19) So, it might be useful to take 33^47 and REWRITE it as (33^28)(33^19) NOTE: later on, we'll apply a nice exponent rule that says "(a^n)(b^n) = (ab)^n"

We get: (32^28)(33^47)(37^19) = (32^28)(33^28)(33^19)(37^19) = (32^28 x 33^28)(33^19 x 37^19) = (32 x 33)^28 (33 x 37)^19 [applied above rule] =(---6)^28 (---1)^19 [I'm focusing solely on the units of each product. So, I use "---" to represent the other digits] =(----6)(----1) [When ----6 is raised to any power the units digit is always 6. The same applies to ----1] = -------6

Re: What is the units digit of the product (32^28) (33^47) (37^19)?
[#permalink]

Show Tags

17 Jul 2018, 21:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________