Prove that 25^11−5^19 is divisible by 31.

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

4 0

Answer:

Divisible by 3 is the answer

Step-by-step explanation:

First get everything to have the same base of 5

25^11 - 5^19

(5^2)^11 - 5^19

5^(2*11) - 5^19

5^22 - 5^19

Now factor out the GCF 5^19 to get

5^22 - 5^19

5^(19+3) - 5^(19+0)

5^19*5^3 - 5^19*5^0

5^19(5^3 - 5^0)

5^19(125 - 1)

5^19*(124)

At this point, we factor the 124 into 31*4 to end up with this full factorization: 5^19*31*4

Therefore, 25^11 - 5^19 is equivalent to 5^19*31*4

Since 31 is a factor of the original expression, this means the original expression is divisible by 31.

You might be interested in

Can someone help me out

Sophie [7]

Firstly, you want to multiply (f^4)^2. $\frac{2*f^8*f^2}{8f^{12}}$

Next, multiply everything on the numerator. $\frac{2f^{10}}{8f^{12}}$

Next, divide. $\frac{2f^{10}}{8f^{12}}=\frac{1}{4} f^{-2}$

Now remember that f^-2 = (1/f^2), so just convert the negative powered exponent to that, and your answer will be $\frac{1}{4f^2}$

6 0

3 years ago

Establish the identity.<br> (2 cos 0-6 sin 0)² + (6 cos 0+2 sin 0)2 = 40

kogti [31]

Rewriting the left-hand side as follows,

$(2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40$