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03 Feb 2005, 20:57
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If 100<n<120, what is the value of n
I. n divided by 4 leaves a remainder of 3.
II. n-2 when divided by 5 doesn't leave any remainder .
Can anyone tell me how to do these stupid remainder , multiplier problems using algebra, picking numbers takes forever and also have a chance of missing something. ??
I. n divided by 4 leaves a remainder of 3.
II. n-2 when divided by 5 doesn't leave any remainder .
Can anyone tell me how to do these stupid remainder , multiplier problems using algebra, picking numbers takes forever and also have a chance of missing something. ??
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03 Feb 2005, 21:28
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(C)
(A) alone doesn't do it because we have 102, 107, 111, 115, and 119.
(B) alone doesn't do it either because we have n-2 as (102 -2) /5 = 20, ......(117-2) / 5 = 23
Now combining both, i think n = 102
(A) alone doesn't do it because we have 102, 107, 111, 115, and 119.
(B) alone doesn't do it either because we have n-2 as (102 -2) /5 = 20, ......(117-2) / 5 = 23
Now combining both, i think n = 102
03 Feb 2005, 22:04
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I got (C) too.
Can't derive everything by algebra though.
(I) n=4k+3 not sufficient for there are more than one bt (100,120)
(II) n=5m+2 not sufficient for there are more than one bt (100,120)
Combine, I got the idea that these two can only have a common number every 20(=4*5) numbers, so would only have one in (100,120). However I can't express it, so I had to write out the numbers.
Ok I just got an idea:
n=4k+3=5m+2
4k=5m-1=5(m-1)+5-1=5(m-1)+4
4(k-1)=5(m-1)
In other words, k-1 must be divisible by 5. k-1=5t
n=4k+3=4(k-1)+4+3=20t+7
You can get that only one number falls into (100,120), which is 107.
This method though, is more time consuming than writing the numbers out. It is nontherless helpful to us now to understand how to use algebra to derive things.
Hopefully I have not confused everybody.
4(k-1)
Can't derive everything by algebra though.
(I) n=4k+3 not sufficient for there are more than one bt (100,120)
(II) n=5m+2 not sufficient for there are more than one bt (100,120)
Combine, I got the idea that these two can only have a common number every 20(=4*5) numbers, so would only have one in (100,120). However I can't express it, so I had to write out the numbers.
Ok I just got an idea:
n=4k+3=5m+2
4k=5m-1=5(m-1)+5-1=5(m-1)+4
4(k-1)=5(m-1)
In other words, k-1 must be divisible by 5. k-1=5t
n=4k+3=4(k-1)+4+3=20t+7
You can get that only one number falls into (100,120), which is 107.
This method though, is more time consuming than writing the numbers out. It is nontherless helpful to us now to understand how to use algebra to derive things.
Hopefully I have not confused everybody.
4(k-1)
