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Updated on: 19 Jan 2014, 09:37
Originally posted by jovialsarthak on 19 Jan 2014, 01:33.
Last edited by Bunuel on 19 Jan 2014, 09:37, edited 2 times in total.
Last edited by Bunuel on 19 Jan 2014, 09:37, edited 2 times in total.
Renamed the topic and edited the question.
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C
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Difficulty:
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Question Stats:
76% (00:59) correct
24%
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based on 212
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What will be the remainder of: 12345678910/8
A. 4
B. 5
C. 6
D. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
A. 4
B. 5
C. 6
D. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
19 Jan 2014, 09:40
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mau5
jovialsarthak
What will be the remainder of:
123........10/8
a. 4
b. 5
c. 6
d. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
123........10/8
a. 4
b. 5
c. 6
d. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
For any # to be divisible by 8, the last 3 digits of that # should be divisible by 8
Given # --> 12345678910 --> \(10^{10}+2*10^9+.... 8000+910\)
All the numbers above are divisible as they all end in 000. Thus, the remainder is what is obtained by\(\frac{910}{8} = 6\)
C.
You can do the same by breaking 12345678910 only into two parts: 12345678910 = 12345678000 + 910.
12345678000 is divisible by 8, because it ends with 000 and the remainder when 910 is divided by 8 is 6.
Answer: C.
General Discussion
19 Jan 2014, 01:41
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jovialsarthak
What will be the remainder of:
123........10/8
a. 4
b. 5
c. 6
d. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
123........10/8
a. 4
b. 5
c. 6
d. 3
PS: Apart from time consuming traditional Division method, is there any trick to attempt such type of questions?
For any # to be divisible by 8, the last 3 digits of that # should be divisible by 8
Given # --> 12345678910 --> \(10^{10}+2*10^9+.... 8000+910\)
All the numbers above are divisible as they all end in 000. Thus, the remainder is what is obtained by\(\frac{910}{8} = 6\)
C.
