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Source: — Data Sufficiency |
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by shankar.ashwin » Sat Nov 19, 2011 9:53 pm
E IMO.

10^(m or n) will have sum of the digits as 1.

Possible remainders when divided by 3, are 0,1 and 2.

Statement 1:

m > n, notice when m=5 and n=2, both expressions are divisible by 3. We can increase m=6, remainder becomes 1. So different possibilities - Insufficient.

Statement 2:

n=6, we dont know 'm' here, cannot be determined.

Even combining, we can have different remainders for the expression. (say n=6 and m=9 will have same remainder, whereas m can take value of 10 when remainder changes)

E IMO
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by mehrasa » Sat Nov 19, 2011 10:07 pm
shankar.ashwin wrote:E IMO.

10^(m or n) will have sum of the digits as 1.

Possible remainders when divided by 3, are 0,1 and 2.

Statement 1:

m > n, notice when m=5 and n=2, both expressions are divisible by 3. We can increase m=6, remainder becomes 1. So different possibilities - Insufficient.

Statement 2:

n=6, we dont know 'm' here, cannot be determined.

Even combining, we can have different remainders for the expression. (say n=6 and m=9 will have same remainder, whereas m can take value of 10 when remainder changes)

E IMO
i also agree with u but I donno why I wrote on my note that the answer is B.. but there is a point u missed for stat 2
when n=6, whatever m is the remainder will be 1
but for 10^6 +m we ca not say bcuz the important figure here (for divisibility by 3) is the number added to 10^whatever
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by pemdas » Sun Nov 20, 2011 3:16 am
the question asks if remainder (n+1)/3 > remainder (m+1)/3?
note: the tens will be divided by 3 with the remainder of 1 added to either n or m.
st(1) m>n, if we divide both sides by 3 and add 1/3 (consider given m,n=positive integers) the sign won't change, hence m/3 +1/3 > n/3 +1/3. We can answer No to the original question, Sufficient.
st(2) n/3=2 translates into n=6, but we don't know anything about m. Not Sufficient.

a

after solving this q. I have noticed the poster to come up with agreement about ans. E, what's the source and is E OA?
mehrasa wrote:if m and n are positive integers. is remainder of [10^m)+n]/3 larger than remainder [(10^n)+m]/3?
1)m>n
2) n/3=2

plz provide ur explanation
thnx
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by mehrasa » Sun Nov 20, 2011 4:59 am
pemdas wrote:the question asks if remainder (n+1)/3 > remainder (m+1)/3?
note: the tens will be divided by 3 with the remainder of 1 added to either n or m.
st(1) m>n, if we divide both sides by 3 and add 1/3 (consider given m,n=positive integers) the sign won't change, hence m/3 +1/3 > n/3 +1/3. We can answer No to the original question, Sufficient.
st(2) n/3=2 translates into n=6, but we don't know anything about m. Not Sufficient.

a

after solving this q. I have noticed the poster to come up with agreement about ans. E, what's the source and is E OA?
mehrasa wrote:if m and n are positive integers. is remainder of [10^m)+n]/3 larger than remainder [(10^n)+m]/3?
1)m>n
2) n/3=2

plz provide ur explanation
thnx
hey pemdas
if u read the Q stem again, it is (10^m)+n/3 and (10^n)+m/3 NOT simply (n+1)+3

thnx
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by LalaB » Sun Nov 20, 2011 5:34 am
mehrasa wrote:if m and n are positive integers. is remainder of [10^m)+n]/3 larger than remainder [(10^n)+m]/3?
1)m>n
2) n/3=2

plz provide ur explanation
thnx
+1 for E

stmnt 1 is insuff, because m could be 3 , n=2 then the remainder of [(10^n)+m]/3 is 0, and the remainder of [10^m)+n]/3 is 2. so [10^m)+n]/3 is larger than [(10^n)+m]/3. but if m=2 , n=1
the situation will change.

stmnt 2 says that n= 6. since we have no info about m, this stmnt is insuff

both stmnt- assume that n=6 (from stmnt 2) and m=7(from stmnt 1 that m must be greater than n), then the remainder of [(10^n)+m]/3 is greater than that of [10^m)+n]/3 .but lets suppose that n=6 and m =9, then the remainders are equal

so ,the answ is E
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by pemdas » Sun Nov 20, 2011 10:35 am
@mehrasa, i've read the question before posting reply - no worries
i hope you don't revise the question's text from original
is remainder of [10^m)+n]/3 larger than remainder [(10^n)+m]/3
to just entered
(10^m)+n/3 and (10^n)+m/3
now the click here is that when i say (n+1)/3 i mean (n+1)/3 which has the same remainder as [(10^m)+n]/3. Please check m=1, positive integer -> (10^1+n)/3 should be equivalent to 3 1/3 +n/3, hence we have remainder 1 (besides the whole part 3) and number n divided by 3.

You can try the fate with m=2,3,4,100,etc. it's all the same matter -> (n+1)/3

Having said about n, i also mean (m+1)/3 for m BUT not
simply (n+1)+3
i hope we are learning from our posts and not burlesquing each other

do me favor, plz reread my post
mehrasa wrote:
pemdas wrote:the question asks if remainder (n+1)/3 > remainder (m+1)/3?
note: the tens will be divided by 3 with the remainder of 1 added to either n or m.
st(1) m>n, if we divide both sides by 3 and add 1/3 (consider given m,n=positive integers) the sign won't change, hence m/3 +1/3 > n/3 +1/3. We can answer No to the original question, Sufficient.
st(2) n/3=2 translates into n=6, but we don't know anything about m. Not Sufficient.

a

after solving this q. I have noticed the poster to come up with agreement about ans. E, what's the source and is E OA?
mehrasa wrote:if m and n are positive integers. is remainder of [10^m)+n]/3 larger than remainder [(10^n)+m]/3?
1)m>n
2) n/3=2

plz provide ur explanation
thnx
hey pemdas
if u read the Q stem again, it is (10^m)+n/3 and (10^n)+m/3 NOT simply (n+1)+3

thnx
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by mehrasa » Sun Nov 20, 2011 11:01 am
thnx pemdas for ur comprehensive answer.. I did not revise my Q stem... if I initially put all in Bracket, becuz I did not want the Q to create ambiguity and it shows that all expression is divided by 3 not for example just n
actually i am not sure about OA... i had written this in my note... let's see what is other people's opinion and maybe we receive an expert opinion on this..

all the best
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by pemdas » Sun Nov 20, 2011 11:53 am
Let me change my answer from A to E finally with the full proof for myself using my own solution in the prior post, i had to test the numbers too, the approach seems sustainable
st(1) m>n, if we divide both sides by 3 and add 1/3 (consider given m,n=positive integers) the sign won't change, hence m/3 +1/3 > n/3 +1/3. We can answer No to the original question, Sufficient.
we input m=4,5,6 (i don't start from 1 as it says m>n and n cannot be 0 for the all numbers are positive integers) as the remainder cycles for the divisor of 3 to test the distance between numbers here and correspondingly assess the remainder:
m=4, n=3 -> 4/3 + 1/3 > 3/3 +1/3 true, the remainder on LHS is 2 and on the RHS is 1
m=5, n=4 -> 5/3 +1/3 >! (should be < ) 4/3 +1/3 true, the remainder on LHS is 0 and on the RHS is 2
m=6,n=5 -> 6/3+1/3 > 5/3 + 1/3 true, the remainder on LHS is 1 and on the RHS is 0

we have Yes/No situation and this will make statement (1) alone Not Sufficient

statement(2) suggests n=6 alone is Not Sufficient

combined statements(1&2) we have n=6 and m>n
(n+1)/3 will always return the remainder of 1 -> (6+1)/3= 2 + 1/3
while m>6 may also return the remainder of 0,1 and 2 Not Sufficient

Now, I'm sure it's E
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