I came across this post in gmatclub.com.
Source:
[spoiler]https://gmatclub.com/forum/collection-of ... 74776.html[/spoiler]
If r is the remainder when the positive integer n is divided by 7, what is the value of r
1. when n is divided by 21, the remainder is an odd number
2. when n is divided by 28, the remainder is 3
The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.
St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.
But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
St 2: when n is divided by 28 the remainder is 3.
As 7 is a factor of 28, the remainder when divided by 7 will be 3
SUFFICIENT
As per the reasoning given in 2nd case,it says that since the remainder when divided by 28 is 3 and since 7 is a factor of 28, therefore the remainder will be 3 for 7 also.
My question here is that if we consider the case: 19 divided by 4
remainder is 3. So 19 when divided by 2 should also give a remainder of 3. Since 2 is a factor 4.
Can somebody explain this??
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Remainder doubts:
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abhishekswamy
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galaxian
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Yes, reminder is 3 in your 2nd case, but since 2 < 3, you again divide 3 by 2 which gives 1 as the reminder. [How can a reminder be greater than the divisor - the original no]
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winniethepooh
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Your reasoning in the first statement is incorrect.
Consider 24. It is divisble by 21 in 1 time and the remainder is 3.
It is also divisible by 7 in 3 times and the remainder is 3.
The only fact that 1 is not sufficient is - as we are asked the precise value of r, we can't be sure about the remainder being either 1, 3, 5 or 9.
Statement 2, sure is sufficient. As remainder of n after being divided by 28 is 3. As seven is a multiple of 28 anything divisible by 28 having a remainder less than 7 will have the same ramainder when divided by 7.
Hope this helps.
Consider 24. It is divisble by 21 in 1 time and the remainder is 3.
It is also divisible by 7 in 3 times and the remainder is 3.
The only fact that 1 is not sufficient is - as we are asked the precise value of r, we can't be sure about the remainder being either 1, 3, 5 or 9.
Statement 2, sure is sufficient. As remainder of n after being divided by 28 is 3. As seven is a multiple of 28 anything divisible by 28 having a remainder less than 7 will have the same ramainder when divided by 7.
Hope this helps.
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Anurag@Gurome
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abhishekswamy wrote:I came across this post in gmatclub.com.
Source:
[spoiler]https://gmatclub.com/forum/collection-of ... 74776.html[/spoiler]
If r is the remainder when the positive integer n is divided by 7, what is the value of r
1. when n is divided by 21, the remainder is an odd number
2. when n is divided by 28, the remainder is 3
The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.
St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.
But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
St 2: when n is divided by 28 the remainder is 3.
As 7 is a factor of 28, the remainder when divided by 7 will be 3
SUFFICIENT
As per the reasoning given in 2nd case,it says that since the remainder when divided by 28 is 3 and since 7 is a factor of 28, therefore the remainder will be 3 for 7 also.
My question here is that if we consider the case: 19 divided by 4
remainder is 3. So 19 when divided by 2 should also give a remainder of 3. Since 2 is a factor 4.
Can somebody explain this??
This should make things clearer.
Since n leaves a remainder of 3 when divided by 28, we can write
n = 28k + 3 where k is an integer.
n/7 = 4k + 3/7.
Since 3 is less than 7, 3/7 leaves a remainder of 3.
This implies that n on being divided by 7 leaves a remainder of 3.
Next, take your case.
19 = 4*4 + 3.
19/2 = 4*2 + 3/2.
Here 3 is more than 2.
Note that any number on being divided by 2 can only give 0 or 1 as remainder.
Also 3/2 = 1 ½.
Remainder is hence 1.
If any number on being divided by 4 would have given a remainder less than 2, then the remainder would have been the same for that number on being divided by 2.
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ArpanaAmishi
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St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.
But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
Someone please explain this in detail ... I am not clear on this...especially on 'it cannot be 7, 3 or 9 '
But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
Someone please explain this in detail ... I am not clear on this...especially on 'it cannot be 7, 3 or 9 '
