Find the greatest number that divides451, 328 and 673 leaving 1, 4 and 1,respectively as remainders.

You're looking for the largest number x such that

x ≡ 1 (mod 451)

x ≡ 4 (mod 328)

x ≡ 1 (mod 673)

Recall that

x ≡ a (mod m)

x ≡ b (mod n)

is solvable only when a ≡ b (mod gcd(m, n)). But this is not the case here; with m = 451 and n = 328, we have gcd(m, n) = 41, and clearly

1 ≡ 4 (mod 41)

is not true.

So there is no such number.

can i see the following ?

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.