Answer
see details below
Work Step by Step
a. To verify 128 ≡ 2 (mod 7), we can use the following algebraic manipulation :
128 − 2 = 126 = 7 * 18 <=> 128 − 2 = 7 * 18
which can be rewritten as: 128 ≡ 2 (mod 7)
Also, to verify 61 ≡ 5 (mod 7), we have :
61 – 5 = 56 = 7 * 8 <=> 61 – 5 = 7 * 8
which can be rewritten as: 61 ≡ 5 ( mod 7 )
b. To verify (128 + 61) ≡ ( 2 + 5) (mod 7), we have :
128 – 2 = 7 * 18
61 – 5 = 7 * 8
Therefore, (128 + 61) − ( 2 + 5 ) = ( 128 – 2 ) + ( 61 – 5 )
= 7 * 18 + 7 * 8
= 7 * 26
=> (128 + 61) − ( 2 + 5 ) = 7 * 26
which can be rewritten as: (128 + 61) ≡ ( 2 + 5) ( mod 7 )
c. To verify (128 − 61) ≡ ( 2 − 5) (mod 7), we have :
128 – 2 = 7 * 18
−61 – ( – 5 ) = −56 = 7 * ( −8 )
Therefore, (128 − 61) − ( 2 − 5 ) = ( 128 – 2 ) + [ −61 – ( – 5 ) ]
= 7 * 18 + 7 *(− 8)
= 7 * 10
=> (128 − 61) − ( 2 − 5 ) = 7 * 10
which can be rewritten as: (128 − 61) ≡ ( 2 − 5) ( mod 7 )
d. To verify (128 * 61) ≡ ( 2 * 5 ) (mod 7), we have :
128 * 61 = 7808 = 7 * 1115 + 3
2 * 5 = 10 = 7 * 1 + 3
So, both 128 * 61 and 2 * 5 leave a remainder of 3 when divided by 7, therefore, we can
say that 128 * 61 and 2 * 5 are congruent modulo 7.
Therefore, ( 128 * 61 ) ≡ ( 2 * 5 ) ( mod 7 )
e. To verify $( 128^2 ) ≡ ( 2^2 ) ( \mod 7 )$, we have :
$128^2 = ( 7 * 18 + 2 )^2 = (7 * 18)^2 + 2 * (7 * 18) * 2 + 2^2$
$ = 7 * 7 * 18 * 18 + 7 * 18 * 4 + 4$
$ = 7 * 2340 + 4$
$ 2^2 = ( 7 * 0 + 2 )^2 = ( 7 * 0 )^2 + 2 * ( 7 * 0 ) * 2 + 4 = 4$
So, both $128^2$ and $2^2$ leave a remainder of 4 when divided by 7, therefore, we can
say that $128^2$ and $2^2$ are congruent modulo 7.
Therefore, $(128^2) ≡ ( 2^2 ) ( \mod 7 )$
