Repetitive products

Algebra Level 2

$A$ is positive and $A\times A \times A = 8\times 8\times 8\times 8$ .

What is $A\times A \times A \times A ?$

16\times16\times16\times16

8\times 8\times 8\times 8\times 8

4\times 4\times 4\times 4\times 4\times 4

4 solutions

Richard Costen
Mar 21, 2017

$A^3=8^4=(2^3)^4=(2^4)^3=16^3$

Since A x A x A = 16 x 16 x 16

A x A x A x A = 16 x 16 x 16 x 16

Rahil Sehgal
Mar 21, 2017

Very neat and clear work! Thanks!

Here's a bonus question for you:

If $A$ is not necessarily a positive number, must $A\times A \times A\times A$ be equal to $16\times 16\times16\times16$ ONLY? Why or why not?

Pi Han Goh - 4 years, 2 months ago

No, $A$ can also have $-16$ as its value as $(-16) \times (-16) \times(-16 )\times(-16 )$ has the same value as $16^{4}$ which is $A^{4}$

Rahil Sehgal - 4 years, 2 months ago

But if $A = -16$ , then $A\times A\times A= 8\times8\times8\times8$ is no longer true.

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – Oh sorry... I didn't notice that... Then $A$ can only have positive value..

Rahil Sehgal - 4 years, 2 months ago

@Rahil Sehgal – But, can $A$ be a non-real number, like say $A = 1 + i$ ? Why or why not?

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – Let, suppose $A= a+ib$

Then $A^{3} = ((2)^{3})^{4}$

This gives us that $a+ib = 16$

Therefore $b = 0$ and $A = a = 16$ This proves that $A$ is positive integer.

Rahil Sehgal - 4 years, 2 months ago

@Rahil Sehgal – No, that's incorrect. The solution for $x^3 = 1$ are $x = 1, \dfrac{1 \pm \sqrt{3} i }2$ . So, can you find the other solutions for $A^3 = 16^3$ ?

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – I have added a note to my solution... Hope that clarifies.. :)

Also please try my set JEE mathematics

Rahil Sehgal - 4 years, 2 months ago

@Rahil Sehgal – Yes! That's great.

Minor nitpicking: It shouldn't be "If $A$ is unreal...", but it should be "If $A$ is not necessarily a real number, then the other solutions are..."

Hope you enjoyed this question~~

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – Okay.. thank you :)

Rahil Sehgal - 4 years, 2 months ago

A Former Brilliant Member
Jun 10, 2018

$A \times A \times A = 8 \times 8 \times 8 \times 8$

$A^3=8^4=4096$

Raising both sides to $\frac{1}{3}$ gives

$A=16$

$A \times A \times A \times A = 16 \times 16 \times 16 \times 16$

Munem Shahriar
Jan 26, 2018

$8^4 = (2^3)^4 = (2^4)^3 = 16^3$

That means $A = 16$ . Hence $A^4 = \boxed{16^4}$

Repetitive products

4 solutions

0 pending reports