If x is an integer such that 3 x + 4 is divisible by 7, is 4 x + 3 also divisible by 7?
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Quite a clean solution! :)
( ( 3 x + 4 ) + ( 4 x + 3 ) ) ( m o d 7 ) = ( 7 x + 7 ) ( m o d 7 ) ≡ 0 ⟹ if 3 x + 4 ( m o d 7 ) ≡ 0 , then 4 x + 3 ( m o d 7 ) ≡ 0
Here we go,
7 ∣ 3 x + 4
So \(7|3x+4-7|)\ also.
Now \(7|3x-3\)
Hence 7 ∣ x − 1 .
Now see , 7 ∣ 4 x − 4
We can write as, 7 ∣ 4 x − 4 + 7
So 7 ∣ 4 x + 3 for all the integers x in which 7 ∣ 3 x + 4 . ( P R O V E D )
By the way, In some previous year,this was to be proved in RMO
Thank you
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We can write 4 x + 3 = 7 x + 7 − ( 3 x + 4 )
Because 7 x + 7 and 3 x + 4 are divisible by 7, we can conclude that 4 x + 3 is divisible by 7