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What is the remainder when 1990990900032 is divided by 32 ?

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stonecold
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What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   20 Jul 2016, 09:31
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E

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91% (00:55) correct 9% (01:43) wrong based on 203 sessions
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What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0
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E
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KarishmaB
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   20 Jul 2016, 21:20
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stonecold wrote:
What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0


For powers of 2 - A number ending in as many 0s as the exponent of 2 will be divisible by that power of 2.

e.g. 2^4 = 16
A number N ending in four 0s will be divisible by 16.

e.g. 2^5 = 32
A number N ending in five 0s will be divisible by 32.

and so on...

1990990900032 = 1990990900000 + 32
1990990900000 ends in five 0s so will be divisible by 32.
32 is divisible by 32.
So 1990990900032 will be divisible by 32. Remainder will be 0.

Answer (E)
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Anigr16
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What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   20 Jul 2016, 21:13
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stonecold wrote:
What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0


Though i was unaware of the divisibility test for 32 but i guessed the pattern!!
divisibility rule for 4- last two digits must be divisible by 4
divisibility rule for 8- last three digits must be divisible by 8

similarly, divisibility rule for 32 - last five digits must be divisible by 32

Hence, Ans E
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   21 Jul 2016, 09:29
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VeritasPrepKarishma wrote:
stonecold wrote:
What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0


For powers of 2 - A number ending in as many 0s as the exponent of 2 will be divisible by that power of 2.

e.g. 2^4 = 16
A number N ending in four 0s will be divisible by 16.

e.g. 2^5 = 32
A number N ending in five 0s will be divisible by 32.

and so on...

1990990900032 = 1990990900000 + 32
1990990900000 ends in five 0s so will be divisible by 32.
32 is divisible by 32.
So 1990990900032 will be divisible by 32. Remainder will be 0.

Answer (E)



Learnt something new today !!
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BrentGMATPrepNow
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   02 Dec 2017, 15:34
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stonecold wrote:
What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0


100,000 = (10)(10)(10)(10)(10)
= (2)(5)(2)(5)(2)(5)(2)(5)(2)(5)
= (2^5)(5^5)
= (32)(5^5)

So,....
1990990900032 = 1990990900000 + 32
= (19909909)(100000) + 32
= (19909909)((32)(5^5)) + 32
= 32[(19909909)((5^5)) + 1]

So, 1990990900032 is a MULTIPLE of 32, which means the remainder is 0

Answer: A

Cheers,
Brent
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   03 Dec 2017, 11:23
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Split 32 as 8*3
Now consider divisibility tests for 8 and 3
Divisibility by 8: last three digits must be divisible by 8. Therefore 032 divisible by 8 true
Divisibility by 3: sum of digits must be divisible by 3; Now it is easy since the number contains many 9's sum=9*5+1+3+2=51 divisible by 3
Thus, when divided by 32-remainder will be zero
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink] New post   26 Feb 2024, 20:45
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Re: What is the remainder when 1990990900032 is divided by 32 ? [#permalink]
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