What is the remainder when 99009900990099000065 is divided by 32

Question

A)zero
B)one
C)two
D)three
E)four

vitaliyGMAT · Accepted Answer

Imagine we have number

N = abcdefgh … vwxyz  where  abc…  are single digits.

Now we need to find remainder when we divide this number by  2^x

In order to do so we need to pay attention only to last  x  digits. If  x=1  --->  2^1  (  z  ), if  x=2  --->  2^2=4  (  yz  ) (divisibility rule for 4 is just a single case) and so on.

Back to our question:

32 = 2^5  and we need to take into account only last 5 digits of 990099009900990 00065 .

Discarding the preceding zeros we have:

65/32 remainder is 1.

Answer B.

acegmat123 · Answer

99009900990099000065 = 99009900990099000064 + 1 = 32q+1

Therefore remainder = 1

B

iamkuberan · Answer

Ans B

thats the answer i got.. even digit answers are ruled out .. only 1 and 3 remain.. ans comes to be 1

broall · Answer

Simply, 99009900990099 000000 =99009900990099 * 10^6 is divisible by 2^6=64.

Hence, the remainder of 990099009900990000 65 /32 = the remainder of 65/32=1.

mvictor · Answer

we could easily eliminate few choices...
it can't be zero, as 32 multiplied by any number would not yield another number with units digit equal to 5.
Moreover, it can't be any EVEN number as a remainder, because odd - even = odd.
C and E are out.

between B and D:
32 * xx2 - then one
32 * xx1 - then three

to be divisible by 32, it must be divisible by 2 four times.
99009900990099000000 is clearly divisible by 32.
64 is divisible by 32 too
therefore, 99009900990099000064 is divisible by 32. therefore, one is the remainder.

Rico355 · Answer

How do you know that 99009900990099000000 is divisible by 32?

Posted from my mobile device

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What is the remainder when 99009900990099000065 is divided by 32

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What is the remainder when 99009900990099000065 is divided by 32

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