Question

Find the smallest value of k when
a)280k is a perfect square
b)882k is a cube

It seems so simple, don't know why I can't solve it

Answer 1

clue:
\[280k=2^2\times 7 \times 5\times 2\times k\]

Answer 2

The answer is a Because look at baru's clue

Answer 3

Baru, that's exactly where I get stuck. I don't know what to do next.

Answer 4

Let me explain
for a perfect square is a number you can square root without any decimals, so with the previous persons explanation, the first 2 is squared but the other 7,5,2 are not squared. thats why you have to multiply them 7x5x2 gives us 70, so the answer for a is 70.

Answer 5

Thanks! For b it's 882k = 2*3*3*7*7
No number seems to multiply itself thrice, what do I do?

Answer 6

if you factorize any perfect cube, it has to look like \((~)^3\times (~)^3\times (~)^3...\)

Answer 7

in this case,
\( k=2^2 \times 3 \times 7 \)

Answer 8

think like this:
i have 2 '7's, i need 3, so i need to multiply one 7
i have 2 '3's, i need 3, so i need to multiply one 3
i have 1 '2', i need 3, so i need to multiply two 2s

7*3*2*2

Answer 9

An example for finding the smallest value of k to make 12k a perfect cube:

\(12=2^2\times 3\)
To make it a perfect cube, we need the pattern:
\(2^3\times 3^3\)
so we need
\(2^2×3k=2^3\times 3^3\)
giving
\(k=2\times 3^2\)
or
perfect cube = \(2^2 \times 3 \times (2\times 3^2) = 2^3\times 3^3=6^3\)

Answer 10

Excellent!