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
remainder
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Source: Beat The GMAT — Data Sufficiency |
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shankar.ashwin
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mehrasa
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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 2shankar.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
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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pemdas
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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?
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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mehrasa
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hey pemdaspemdas 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
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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LalaB
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+1 for Emehrasa 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
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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pemdas
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@mehrasa, i've read the question before posting reply - no worries
i hope you don't revise the question's text from original
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
do me favor, plz reread my post
i hope you don't revise the question's text from original
to just enteredis remainder of [10^m)+n]/3 larger than remainder [(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.(10^m)+n/3 and (10^n)+m/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
i hope we are learning from our posts and not burlesquing each othersimply (n+1)+3
do me favor, plz reread my post
mehrasa wrote:hey pemdaspemdas 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
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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mehrasa
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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
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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pemdas
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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
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
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: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.
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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