What is the remainder when 1990990900032 is divided by 32 ?

Question

What is the remainder when 1990990900032 is divided by 32 ?
A) 16
B) 8
C) 4
D) 2
E) 0

KarishmaB · Accepted Answer

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)

Anigr16 · Answer

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

CounterSniper · Answer

Learnt something new today !! :-D

BrentGMATPrepNow · Answer

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  (5^5) ) + 1]

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

Answer: A

Cheers,
Brent

jasmine9 · Answer

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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What is the remainder when 1990990900032 is divided by 32 ?

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What is the remainder when 1990990900032 is divided by 32 ?

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