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by arjunshn » Wed Apr 13, 2011 4:34 am
Nine different numbers are selected randomly from the integers 350 to 400 ,inclusive, and each number is divided by 9.What is the sum of the remainders?
(1) The range of the nine remainders is 8.
(2) The nine numbers selected are consecutive integers.
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Source: — Data Sufficiency |
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by bubbliiiiiiii » Wed Apr 13, 2011 4:43 am
IMO E!
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Pranay
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by arjunshn » Wed Apr 13, 2011 4:50 am
hey i just realised any 9 consective numbers in that interval when divided by 9 ,their remainders amt to the same sum
answer is b
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by HSPA » Wed Apr 13, 2011 6:04 am
Okay, there is no difference btw the options A and B
So we are left with two options D/E

Do for 360 to 368 and also do for 362 to 370

there result can vary as remainder can vary by value of 1

I agree with Pranay...
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by arjunshn » Wed Apr 13, 2011 6:11 am
from 360 to 368 u get 0,1,2,3,4,5,6,7,8 remainders
from 362 to 370 u get 2,3,4,5,6,7,8,0,1 remainders
so there sum is the same
hence b
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by HSPA » Wed Apr 13, 2011 6:13 am
Do you mean D? B is a typo on your side i guess
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by arjunshn » Wed Apr 13, 2011 6:17 am
i mean the answer is B..because you pick any nine consective numbers from 350 to 400 inclusive..they have same remainders thus same sum..hence B
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by HSPA » Wed Apr 13, 2011 6:34 am
arjunshn wrote:i mean the answer is B..because you pick any nine consective numbers from 350 to 400 inclusive..they have same remainders thus same sum..hence B
B means statement two alone is sufficient to answer the question and A is wrong
but " A aur B donno ek hi option hey"

In reality we are left with D/E choices only... the choices A,B,C are not valid.. because both optins are SAME

Hope this helps...
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by arjunshn » Wed Apr 13, 2011 6:35 am
is this correct?
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by Brent@GMATPrepNow » Wed Apr 13, 2011 6:37 am
arjunshn wrote:Nine different numbers are selected randomly from the integers 350 to 400 ,inclusive, and each number is divided by 9.What is the sum of the remainders?
(1) The range of the nine remainders is 8.
(2) The nine numbers selected are consecutive integers.
Statement 1
If the range of the nine remainders is 8, then the numbers could be:
351 (Remainder0), 360 (R0), 369 (R0), 378 (R0), 387 (R0), 396 (R0), 368 (R8), 361 (R1), 362 (R2), in which case the sum of the remainders (after dividing by 9) is 11
However, if the range of the nine remainders is 8, then the numbers could also be:
351 (Remainder0), 360 (R0), 369 (R0), 378 (R0), 387 (R0), 396 (R0), 368 (R8), 361 (R1), 363 (R3), in which case the sum of the remainders (after dividing by 9) is 12
Since statement 1 yields at least two different answers to the target question, it is NOT sufficient

Statement 2
When it comes to consecutive integers, there is a nice rule that says "If there are n consecutive integers, then exactly one of those integers is divisible by n"

So, of the nine consecutive numbers, exactly one of them must be divisible by 9. In other words, one of them must have a remainder of 0 when divided by 9.
Since the remaining 8 numbers are consecutive, one must have a remainder of 1 when divided by 9.
Another must have a remainder of 2 when divided by 9
And so on.

So, the sum of the remainder (when divided by 9) MUST be 0+1+2+3+4+5+6+7+8=36

This means statement 2 is sufficient and the answer is B
Brent Hanneson - Creator of GMATPrepNow.com
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by HSPA » Wed Apr 13, 2011 6:47 am
Bernt,
Thanks for the reply... i took range = maximum value - minimum value = 396-351(for items in blue text)

sorry i missed range of remainders... i took range of numbers...

enough math for one day..I guess for me... but enjoyed the discussion..
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by MAAJ » Thu Apr 14, 2011 3:38 pm
Just adding:

The sum of "k" CI is divisible by "k" if "k" = odd; On the other hand if "k" = even, then the sum of "k" CI is never divisible by "k"

(1+2+3)/3 = 6/3 = 2
(1+2+3+4)/4 = 10/4 = 2.5
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