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
Stacey Warren - Expert brainly.com
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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.
Thanks! For b it's 882k = 2*3*3*7*7
No number seems to multiply itself thrice, what do I do?
if you factorize any perfect cube, it has to look like \((~)^3\times (~)^3\times (~)^3...\)
in this case,
\( k=2^2 \times 3 \times 7 \)
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
An example for finding the smallest value of k to make 12k a perfect cube:
To make it a perfect cube, we need the pattern:
so we need
perfect cube = \(2^2 \times 3 \times (2\times 3^2) = 2^3\times 3^3=6^3\)