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Arvind42
Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6

Units digit of the bases are repeat in the following cycle - 4 for every 2nd time, 9 for every 2nd time, 3 for every 4th time, 7 for every 4th time, 2 for every 4th time.

the units digit of the expression will be - 4*9*3*1*2 whose unit digit will be 6 IMO option E

Arvind42
see highlighted part ; 9^15 ; shouldnt it be 1 ?
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Arvind42
Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6

Units digit of the bases are repeat in the following cycle - 4 for every 2nd time, 9 for every 2nd time, 3 for every 4th time, 7 for every 4th time, 2 for every 4th time.

the units digit of the expression will be - 4*9*3*1*2 whose unit digit will be 6 IMO option E

Arvind42
see highlighted part ; 9^15 ; shouldnt it be 1 ?

Archit3110 9 repeats in a cycle of 2 i.e. 9 and 1 only. So for all odd powers units digit will be 9 and even powers 1. So i took 9.
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my bad ;; i did it right for others but got it wrong for 9 .. :roll:

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Arvind42
Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6

Units digit of the bases are repeat in the following cycle - 4 for every 2nd time, 9 for every 2nd time, 3 for every 4th time, 7 for every 4th time, 2 for every 4th time.

the units digit of the expression will be - 4*9*3*1*2 whose unit digit will be 6 IMO option E

Arvind42
see highlighted part ; 9^15 ; shouldnt it be 1 ?

9^15 = 205891132094649

I also ended up at AC E with 6.
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Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6
\(4^{93}*9^{15}*3^{81}*7^{56}*2^9\)

= \(2^{186}*3^{30}*3^{81}*7^{56}*2^9\)

= \(2^{195}*3^{111}*7^{56}\)

Now, \(2^{195} = 2^{4*48+3}\) ; So this will have units digit 8

\(3^{111} = 3^ {4*36 + 3}\) will have units digit as 7

\(7^{56} = 7^{4*24}\) will have units digit as 1

Finally we have units digit as 8*7*1 = xx56 , thus Answer must be (E) 6
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Archit3110
Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6

use cyclicity
we get 4*9*3*1*2 ; 216
unit 6
IMO E

Can you explain the cyclicality? howd you know to pick 3 instead of another number or the 1?
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Bunuel
What is the units digit of \(4^{93}*9^{15}*3^{81}*7^{56}*2^9\) ?

A. 0
B. 1
C. 4
D. 5
E. 6


The pattern for a base of 4 is: 4-6-4-6, so we see that when 4 is raised to an odd power, the units digit is 4. Thus, the units digit of 4^93 is 4.

The pattern for a base of 9 is: 9-1-9-1, so we see that when 9 is raised to an odd power, the units digit is 9. Thus, the units digit of 9^15 is 9.

The pattern for a base of 3 is: 3-9-7-1-3-9-7-1, so we see that when 3 is raised to a power that is a multiple of 4, the units digit is 1. Thus, 3^80 has a units digit of 1, so 3^81 has a units digit of 3.

The pattern for a base of 7 is: 7-9-3-1-7-9-3-1, so we see that when 7 is raised to a power that is a multiple of 4, the units digit is 1. Thus, the units digit of 7^56 is 1.

Lastly, since 2^9 is a pretty small number, we see that 2^9 = 512, so it has a units digit of 2.

Since 4 x 9 x 3 x 1 x 2 = 216, the units digit of the original product is 6.

Answer: E
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