Flipping coefficients

If x x is an integer such that 3 x + 4 3x+4 is divisible by 7, is 4 x + 3 4x+3 also divisible by 7?

Yes, always No, never Yes, but not all the time

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3 solutions

Skye Rzym
Apr 9, 2017

We can write 4 x + 3 = 7 x + 7 ( 3 x + 4 ) 4x+3 = 7x+7-(3x+4)

Because 7 x + 7 7x+7 and 3 x + 4 3x+4 are divisible by 7, we can conclude that 4 x + 3 is divisible by 7 \boxed{4x+3 \text{ is divisible by } 7}

Quite a clean solution! :)

toshali mohapatra - 4 years, 1 month ago

( ( 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 \left((3x+4)+(4x+3)\right)\pmod {7}=(7x+7)\pmod {7}\equiv 0\\ \implies \text{if } 3x+4\pmod{7}\equiv 0,\\\text{then } 4x+3\pmod{7}\equiv 0

Md Zuhair
Apr 9, 2017

Here we go,

7 3 x + 4 7|3x+4

So \(7|3x+4-7|)\ also.

Now \(7|3x-3\)

Hence 7 x 1 7|x-1 .

Now see , 7 4 x 4 7|4x-4

We can write as, 7 4 x 4 + 7 7|4x-4+7

So 7 4 x + 3 7|4x+3 for all the integers x x in which 7 3 x + 4 7|3x+4 . ( P R O V E D ) \Bigg(\boxed{PROVED}\Bigg)

By the way, In some previous year,this was to be proved in RMO

Thank you

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