If R is the remainder of the expression (10^5 +1)(10^8 +3)/4 then 4R =
A. 0
B. 4
C. 8
D. 12
E. 16
A. 0
B. 4
C. 8
D. 12
E. 16
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if db=3, dividing it by 4 will leave a reminder of 2, so 4R = 4*2=8p2pg wrote:IMO D
(a+b)(c+d) = (10^5+1)(10^8+3)
Here since any multiple of 100 will be divisible by 4, only db needs to be tested for divisibility. Here db = 3. So R=3.
Nope, on second look I think I made the mistake. The product of the 2 numbers will yield 3 in the units digit. When the product is divided by 4, the last digit will be the reminder. So, the answer should be 12.PAB2706 wrote:vemuri pls elaborate...
i got R=3
4R=12.
i guess i am missing something.
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