If p = 1008^m*998^n, where m and n are positive integers, what will be

Question

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

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dimdiak · Answer

8=2^3. So (2^3m)(2^3n)=> 2^3(m+n)

1) If m=2. 2^6 x 2^3m NS

2) If m+n=13. 2^3(13) Suff

B

Kinshook · Answer

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

Unit digit of 8ˆ1 = 8
Unit digit of 8ˆ2 = 4
Unit digit of 8ˆ3 = 2
Unit digit of 8ˆ4 = 6
Unit digit of 8ˆ5 = 8
Unit digit of 8ˆ6 = 4
Unit digit of 8ˆ7 = 2
Unit digit of 8ˆ8 = 6

Cyclicality of unit digit of 8ˆn = {8,4,2,6,....}

Unit digit of  p=(1008)^m * (998)^n  = Unit digit of  8ˆ{m+n}

(1) m = 2
Since n is unknown
NOT SUFFICIENT

(2) m + n = 13
Unit digit of  p=(1008)^m * (998)^n  = Unit digit of  8ˆ{m+n} = Unit digit of  8ˆ13 = 8  
SUFFICIENT

IMO B

GCMEMBER · Answer

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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CEdward · Answer

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?

yashikaaggarwal · Answer

Yup, the approach would have changed if the unit digit of the numbers raises to the power m and n were different.

exc4libur · Answer

8^(m+n)=?

(1) insufic
(2) sufic

(B)

HASTOWINGMAT · Answer

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.

bumpbot · Answer

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If p = 1008^m*998^n, where m and n are positive integers, what will be

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