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 339 times ]

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