Question

which smallest number should divide 1188 so that the quotient is perfect cube? ​

Answer 1

Answer:

I think the answer is

one 1

Answer 2

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Solution:

1188 = 2 × 2 × 3 × 3 × 3 × 11

The primes 2 and 11 do not appear in groups of three.

So, 1188 is not a perfect cube. In

the factorisation of 1188 the prime 2 appears only two times and the prime Il appears once.

So, if we divide 1188 by 2 x 2 x 1l = 44. then the prime factorisation of the

quotient will not contain 2 and 11.

Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44.

$the \: rusulting \:perfect \: cube \: is$

$1188 \div 44 = 27( = 3 {}^{3} )1188÷44=27(=3 </p><p>3</p><p> )$