More Remainder

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.

p2pg 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.

if db=3, dividing it by 4 will leave a reminder of 2, so 4R = 4*2=8

Answer should be C

vemuri pls elaborate...


i got R=3

4R=12.

i guess i am missing something.

PAB2706 wrote:vemuri pls elaborate...


i got R=3

4R=12.

i guess i am missing something.

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.

Thanks for pointing out.

There are only 4 remainders possible when dividing a number by 4: 1/4, 2/4, 3/4, or 0.

Taking these remainders and multiplying by 4 equals 4R:

4(1/4) = 1
4(2/4) = 2
4(3/4) = 3
4(0) = 0

The only answer that matches is 0, A.

Am I missing something? What's the OA?

The remainder from a division by 4 can be only 0, 1, 2, and 3. These possible remainders, when multiplied by 4, give us the following possible answers 0, 4, 8, and 12. We can automatically reject e) 16.

As somebody already said (10^5 +1)(10^8 +3)=10^13+10^8+3*10^5+3
Each of the powers of ten divides by 4 with a remainder of 0. However, 3 mod 4 = 3.

Therefore the remainder from the whole operation is 3 and the correct answer is 12.

Bst,
Kal

All the powers of 10 above 10^2 is divisible by 4 hence the remainder is 3

the 4R = 12

numbers of the form (4k+1)(4k+3)
=4K^2+12k+4k+3
remainder is 3
4*3=12