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any number when divided by 9, has a remainder equal to the

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New post 20 Feb 2005, 14:15
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any number when divided by 9, has a remainder equal to the sum of its digits. if the last digit of a seven digit number is 4 and the remaining six digits add up to 23, then the original number must be

a divisible by 7
b divisible by 18
c divisible by 23
d divisible by 27
e a prime number
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New post 20 Feb 2005, 19:56
B for me too

its divisible by 9... sum is 27
and its divisible by 2.. . ends with a 4

divisible by 18
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New post 20 Feb 2005, 20:43
Why do we need this first sentence in the question? :?

“any number when divided by 9, has a remainder equal to the sum of its digitsâ€
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New post 22 Feb 2005, 06:11
so the rule is
the sum of the digits of a multiple of 9 is 9?
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New post 22 Feb 2005, 18:41
thearch wrote:
so the rule is
the sum of the digits of a multiple of 9 is 9?



yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :P
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New post 23 Feb 2005, 09:43
can anyone tell me what is this sentence for "any number when divided by 9, has a remainder equal to the sum of its digits"
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New post 23 Feb 2005, 10:06
ETS doesn't assume you have this knowledge, and doesn't require you to have this knowledge. For people who don't know this rule, this information is required to solve the problem.
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New post 23 Feb 2005, 11:18
HongHu wrote:
ETS doesn't assume you have this knowledge, and doesn't require you to have this knowledge. For people who don't know this rule, this information is required to solve the problem.


ok, then is this rule saying that when 9 divided by 9 leaves a remainder of 9 ? It is talking abt "any" number. What abt non integer numbers ? Doesn't seem to be correct rule. :?
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New post 23 Feb 2005, 11:25
Yes, it is not formulated correctly, I would say. What if the sum of the digits are 14? You need to sum it again until the sum is less than 9.

Also, when we talk about reminders, we are usually talking about integers.
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New post 23 Feb 2005, 12:01
OK I admit I have no clue about this topic. So I strategically guessed and picked 18.

here is why....maybe it can help those who dont know these rules..(me :))


well the last digit is 4 (which means its even, i.e divisible by 2)

the sum of the other remaining digits is 23...

hmm, all I know is that for a number to be divisible by 3, its sum must be divisible by 3... (so we have 23+4=27, which is divisible by 3!)

now, I know that this number cannot be a prime (because of the 4!)eliminate choice (e). Next lets look at 27...since 27 is the sum of the 4+23 above I guess its just a trap answer so eliminate (d). now look at 23, 23 is just one of the numbers from the question stem, eliminate it (C)...now we have B, 18, 18 is a product of 2 and 3. meets both criterias, therefore the likely answer...pick B.


Again I wont do this if I knew how to solve the problem, but I am guessing so why not put some thought into it!
  [#permalink] 23 Feb 2005, 12:01
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