if k and n are integers, is n divisible by 7?
1. n-3=2k
2. 2k-4 is divisible by 7
thank you in advance for your help
OG Extra Math review book
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Statement 1:
n-3=2k
-> n=2k+3
if k = 2, n = 7 YES
if k = 3, n = 9 NO
Insufficient
Statement 2:
2k-4 is divisible by 7
Insufficient since we don't learn anything about n.
Together:
n=2k+3 and 2k-4 is divisible by 7
Now see that when we add 7 to 2k-4, we get 2k+3.
So 2k-4, which is divisible by 7, plus 7, which will also be divisible by 7, is 2k+3.
Since n = 2k+3, it will be divisible by 7.
Sufficient.
(Example: If 2k-4 = 14, then 2k+3=21 and both are divisible by 7.)
n-3=2k
-> n=2k+3
if k = 2, n = 7 YES
if k = 3, n = 9 NO
Insufficient
Statement 2:
2k-4 is divisible by 7
Insufficient since we don't learn anything about n.
Together:
n=2k+3 and 2k-4 is divisible by 7
Now see that when we add 7 to 2k-4, we get 2k+3.
So 2k-4, which is divisible by 7, plus 7, which will also be divisible by 7, is 2k+3.
Since n = 2k+3, it will be divisible by 7.
Sufficient.
(Example: If 2k-4 = 14, then 2k+3=21 and both are divisible by 7.)
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THANK YOU! I finally understand the way you explained it! The book had similar explanation but I couldn't figure out what they were trying to say.tmmyc wrote:Statement 1:
Now see that when we add 7 to 2k-4, we get 2k+3.
So 2k-4, which is divisible by 7, plus 7, which will also be divisible by 7, is 2k+3.
Since n = 2k+3, it will be divisible by 7.
Sufficient.
(Example: If 2k-4 = 14, then 2k+3=21 and both are divisible by 7.)
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We know that individually each is insufficient.
Consider both statements.
n = 2k + 3 --------- (1)
2k - 4 is divisible by 7
2k - 4 = 7a (for some integer a)
=> 2k = 7a +4 ------------------(2)
substitute 2 in 1
n = 7a +4 +3
n= 7a +7
n = 7(a+1) a+1 is another integer say b.
n = 7b
This implies n is divisible by 7. So C.
Consider both statements.
n = 2k + 3 --------- (1)
2k - 4 is divisible by 7
2k - 4 = 7a (for some integer a)
=> 2k = 7a +4 ------------------(2)
substitute 2 in 1
n = 7a +4 +3
n= 7a +7
n = 7(a+1) a+1 is another integer say b.
n = 7b
This implies n is divisible by 7. So C.
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IMO C:
n= 2k + 3
n= 5 , 7, 11,
Not suff
2. 2k-4 = n
n = 0, 2, 4, ..
Not suff
But from 1 and 2
n = 7( m+1)
Hence n multiple of 7
n= 2k + 3
n= 5 , 7, 11,
Not suff
2. 2k-4 = n
n = 0, 2, 4, ..
Not suff
But from 1 and 2
n = 7( m+1)
Hence n multiple of 7
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