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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  [#permalink]

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New post 30 Nov 2016, 18:09
1
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

87% (00:43) correct 13% (01:58) wrong based on 75 sessions

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Re: What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 30 Nov 2016, 20:00
stonecold wrote:
What is the remainder when 99009900990099000065 is divided by 32

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



99009900990099000065 = 99009900990099000064 + 1 = 32q+1

Therefore remainder = 1

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Re: What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 30 Nov 2016, 20:07
Ans B

thats the answer i got.. even digit answers are ruled out .. only 1 and 3 remain.. ans comes to be 1
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Re: What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 03 Dec 2016, 05:37
2
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stonecold wrote:
What is the remainder when 99009900990099000065 is divided by 32

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


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 99009900990099000065.

Discarding the preceding zeros we have:

65/32 remainder is 1.

Answer B.
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Re: What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 03 Dec 2016, 10:49
stonecold wrote:
What is the remainder when 99009900990099000065 is divided by 32

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



Simply, 99009900990099000000=99009900990099 * 10^6 is divisible by 2^6=64.

Hence, the remainder of 99009900990099000065/32 = the remainder of 65/32=1.
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What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 03 Dec 2016, 11:03
stonecold wrote:
What is the remainder when 99009900990099000065 is divided by 32

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


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.
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Re: What is the remainder when 99009900990099000065 is divided by 32  [#permalink]

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New post 27 Sep 2018, 22:31
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Re: What is the remainder when 99009900990099000065 is divided by 32 &nbs [#permalink] 27 Sep 2018, 22:31
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