12 Days of Christmas GMAT Competition - Day 1: If P is a positive inte

last value of P^604 can be determined by using cyclicity
since P is raised to power of even number so possible values of last digit can be 0,1,5,6 ; 0 is possible when P is factor of 10 eg ( 10,20... ) for rest 1 is possible for all odd numbers except 5 , 5 is possible for 5 only and 6 is possible for all even numbers

#1 P is not a multiple of 5.
in that case the last digit can be 1 or 6
insufficient
#2
P is even
last digit can be 1 or 5
from 1 &2
last digit is 6
option C

1) if P = 1, last digit = 1; if P = 6, last digit = 6 => not sufficient

2) if P = 10, last digit = 0; if P = 6, last digit = 6 => not sufficient

(Here I choose easy numbers, with only 1 possible last digit when multiplied by themselves).

Combining both statements, knowing that the last digit will equal the last digit of the power cycle * the number:

cyclicity of 2: 2, 4, 8, 6
cyclicity of 4: 4, 6, 4, 6
cyclicity of 6: 6, 6, 6, 6
cyclicity of 8: 8, 4, 2, 6
10 is a multiple of 5 so I don't consider it.

The cyclicity of all even numbers (except for 10) is 1, 2, or 4 => for P^604, the cycle will always end on 6.
Correct answer C.

If P is a positive integer, what is the last digit in the expanded value of \(P^604\)?

(1) P is not a multiple of 5.
(2) P is even.

Statement 1:

P is not multiple of 5

So, P =2 => 2^604 => Since 604 is divisible by 4, so the last digit of P^604 will be the same as the last digit of P^4

=>Last digit of 2^4 = 6
=> Last digit of 3^4 = 1

Not sufficient

Statement 2

P is even

So, if P = 2 then last digit of 2^4 = 6

But, if P = 10 then last digit of 10^4 = 0

Not sufficient

Statement 1 + 2

If P is not a multiple of 5 and it is even then P = 2*x
=> the last digit of x can be 1, 3, 7, 9
=> P^4 = 2^4 * x^4

2^4 will give last digit 6 and 1^4, 3^4, 7^4, and 9^4 will give last digit 1

So, we can definitely say that the last digit of P^4 = 6

Hence, statement 1 and Statement 2 combined are sufficient.

Hence OA should be C.

OA is C
as A and B are not sufficient alone to answer the question but by combining one can easily answer the question.

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