(a) Start by squaring both sides of the congruence 34 = 81 (mod 100) to prove that 38 = 61 (mod 100) and then prove that 3¹6 = 21 (mod 100). What does this tell you about the last two digits in the decimal repre- sentation of 316? (b) Use the two congruences in Part (24a) and laws of exponents to deter- mine r where 320 = r (mod 100) and r € Z with 0 < r < 100. What does this tell you about the last two digits in the decimal representation of 320? 158 @080 BY NC SA Chapter 3. Constructing and Writing Proofs in Mathematics (c) Determine the last two digits in the decimal representation of 3400 (d) Determine the last two digits in the decimal representation of 4804. Hint: One way is to determine the "mod 100 values” for 4², 44, 48, 416, 432, 464, and so on. Then use these values and laws of exponents to determine r, where 4804 = r (mod 100) and r € Z with 0 < r < 100. (e) Determine the last two digits in the decimal representation of 33356 (f) Determine the last two digits in the decimal representation of 7403

please don't  provide hand wrtten solution....

Why does the linear congruence 25x ≡ 30 (mod 75) have no solutions?

In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?