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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
What is the units digit of the product
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04 Aug 2019, 11:23
04 Aug 2019, 11:23
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Question Stats:
56% (01:45) correct
43% (01:47) wrong based on 67 sessions
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What is the units digit of the product \((32^{28})(33^{47})(37^{19})\)?
A) 0
B) 2
C) 4
D) 6
E) 8
Hint:
A) 0
B) 2
C) 4
D) 6
E) 8
Hint:
Show: ::
There’s a fast approach and a slow approach
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: What is the units digit of the product
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04 Aug 2019, 12:56
04 Aug 2019, 12:56
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GreenlightTestPrep 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
Hint:
A) 0
B) 2
C) 4
D) 6
E) 8
Hint:
Show: ::
There’s a fast approach and a slow approach
-----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
Answer: D
Cheers,
Brent
Re: What is the units digit of the product
[#permalink]
05 Aug 2019, 06:45
05 Aug 2019, 06:45
4
GreenlightTestPrep 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
Hint:
A) 0
B) 2
C) 4
D) 6
E) 8
Hint:
Show: ::
There’s a fast approach and a slow approach
Here,
Lets take the sequence of the unit digit that repeat itself:: Plz see the attach diag
The last digit of power of 2 repeat in a cycle of numbers – 2, 4, 8, 6
The last digit of power of 3 repeat in a cycle of numbers – 3, 9, 7, 1
The last digit of power of 4 repeat in a cycle of numbers – 4, 6
The last digit of power of 7 repeat in a cycle of numbers – 7 , 9 ,3 ,1
The last digit of power of 8 repeat in a cycle of numbers – 8, 4, 2, 6
The last digit of power of 9 repeat in a cycle of numbers – 9,1
Now to the ques.
\((32^{28})(33^{47})(37^{19})\)
\(32^{28}\) = last digit is 2 and to the power of 28. Since \(2^{power}\) repeats after every 4th , hence we divide the power by 4 i.e \(\frac{28}{4}\)= 7 and reminder =0 . The last digit of \(32^{28}\) = 6
Similarly for \(33^{47}\) = here \(3^{power}\) repeats after every 4th, so \(\frac{47}{4}\)= 11 and reminder 3, The last digit will be = 7
And for \(37^{19}\) = here \(7^{power}\) repeats after every 4th, so \(\frac{19}{4}\) = 4 and reminder 3, The last digit will be = 3
Combining the digit = \(6*7*3 = 126\) ,
the last digit for the whole equation is = 6
Attachments
Unit digit.png [ 23.51 KiB | Viewed 6507 times ]
