The value of x is 0 because 4*8 is 32 +6 =38 and 4 * 0 is just 0
Use the Euclidean algorithm to express 1 as a linear combination of
and
.
a.
because
77 = 1*52 + 25
52 = 2*25 + 2
25 = 12*2 + 1
so we can write
1 = 25 - 12*2 = 25*25 - 12*52 = (77 - 52)(77 - 52) - 12*52 = 77^2 - 2*52*77 + 52^2 - 12*52
Taken modulo 77 leaves us with

b. First,
, so really we're looking for the inverse of 25 mod 52. We've basically done the work in part (a) already:
1 = 25*25 - 12*52
Taken modulo 52, we're left with

c. The EA gives
71 = 1*53 + 18
53 = 2*18 + 17
18 = 1*17 + 1
so we get
1 = 18 - 17 = 3*18 - 53 = 3*71 - 4*53
so that taken module 71, we find

d. Same process as with (b). First we have
, and we've already shown that
1 = 3*18 - 53
which means, taken modulo 53, that

The equation would have to be set up as 4b - 4 = 10.36. This is because you’re buying 4 boxes of fruit loops, and $1 taken off each box is $4 total for all four boxes.
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There are 10 marbles in the bag all together.
(8/10) = 80% of the marbles in the bag are white.
So if you pull a marble out of the bag, you have an 80% chance
of pulling a white one.
And if you do it many times, you expect that you'll pull
a white one out about 80% of the times.
80% pf 45 times = 0.8 x 45 = 36 times .