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Bunuel
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If p = 1008^m*998^n where m and n are positive integers, what is the units digit of p?

Units digit of p = Units digit of 8^(m+n)

(1) m = 2
n can take any value
Not Sufficient

(2) m + n = 13
Units digit of p = units digit of 8^13 = 8

Sufficient

Option B

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Is B correct only b/c the base of both exponents share the same unit digit of 8? Otherwise if they were different the approach everyone is using above is incorrect?
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CEdward
Is B correct only b/c the base of both exponents share the same unit digit of 8? Otherwise if they were different the approach everyone is using above is incorrect?
Yup, the approach would have changed if the unit digit of the numbers raises to the power m and n were different.
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Bunuel
If \(p=(1008)^m * (998)^n\), where m and n are positive integers, what is the units digit of p?

(1) m = 2

(2) m + n = 13


8^(m+n)=?

(1) insufic
(2) sufic

(B)
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Option B is correct answer.

When the m=2; possible unit digits can be 2,6,4 for n=1,2,3 respectively.
When m+n=13; the unit digit 8 is multiplied 13 times and this gives us the answer.
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