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Bunuel
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Bunuel
Arrange the following terms in the increasing order of their units digits:

I. 7^5
II. 8^6
III. 12^3

A. I < II < III
B. I < III < II
C. II < III < I
D. II < I < III
E. III < I < II

use cyclicity to determine the last digit
7^5 = 7
8^6 = 4
12^3 = 8
II < I < III option d
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Bunuel Shouldn't the answer be Option E?

Obviously 1 < 2

Because base of 2 > 1 and power of 2 > 1

Then 1 > 3

Because rounding off 7^2 * 7^2 as 50*50 would give 2500 --> 2500*7 gives us 17500 approx.
12*12 = 144 and then 144*12 is somewhere around 1440+ approx (if we consider 144*10) which is way off

Bunuel
Arrange the following terms in the increasing order of their units digits:

I. 7^5
II. 8^6
III. 12^3

A. I < II < III
B. I < III < II
C. II < III < I
D. II < I < III
E. III < I < II
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Anuramaa
Bunuel Shouldn't the answer be Option E?

Obviously 1 < 2

Because base of 2 > 1 and power of 2 > 1

Then 1 > 3

Because rounding off 7^2 * 7^2 as 50*50 would give 2500 --> 2500*7 gives us 17500 approx.
12*12 = 144 and then 144*12 is somewhere around 1440+ approx (if we consider 144*10) which is way off

Bunuel
Arrange the following terms in the increasing order of their units digits:

I. 7^5
II. 8^6
III. 12^3

A. I < II < III
B. I < III < II
C. II < III < I
D. II < I < III
E. III < I < II

It seems that you've missed both to read the question carefully and review the discussion.

Arrange the following terms in the increasing order of their units digits:
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