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:
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 5565 times ]
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