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Bunuel
If x is a positive integer, does the remainder, when \((7^x + 1)\) is divided by 100, have 0 as the units’ digit?

(1) x = 4n + 2, where n is a positive integer.
(2) x > 5



7^x+1=100q + m

Now m will have a 0 as its unit digit when 7^x+1 will end in 0 (10, 20, 30.....)

Also 7^ something can have different digit units since it has a ciclicity=4

7
9
3
1
7

These are the units digits of the resulting numbers from the application of powers to number 7.

Statement one gives us the following values for x---> 2, 6, 10, 14
As you can see they are evenly spaced and if plugged in the original 7^x the result will give always a number ending in 9. Hence 7^x+1 will always end up having a unit digit = 0.
Hence statement 1 is sufficient.

Statement 2 on the other end gives us multiple remainders for the original equation. Hence not sufficient.


Option A
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