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Re: What is the units digit of the product (32^28) (33^47) (37^19)?
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10 Apr 2017, 08:55
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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)?
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09 Apr 2017, 09:24
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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)?
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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)?
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04 Aug 2019, 11:55
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