Math, asked by seenabennyseenabenny, 5 months ago

is 1188 a perfect cube? if not by which smallest natural number should 1188 be divided so that the quotient is a perfect cube​

Answers

Answered by Anonymous
24

Answer:

Hence, the smallest number by which 1188 must be divided to obtain a perfect cube is 2² ×11=44.

Step-by-step explanation:

 \sf \: Prime \:  factorising \:  1188, we  \: get,

 \sf \: 1188 \: = \: 2×2×3×3×3×11

 \implies22×33×11.

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 2's is 2, number of 3's is 3 and number of 11's is 1.

So we need to divide 2² and 11 from the factorization to make 1188 a perfect cube.

Hence, the smallest number by which 1188 must be divided to obtain a perfect cube is 2² ×11

 \huge {\boxed { \boxed{ \underline{ \green44}}}}

Answered by h2556deepak
2

Answer:

Prime factorising 1188, we get,

1188=2×2×3×3×3×11

=2^2x 3^3 x 11

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 2's is 2, number of 3's is 3 and number of 11's is 1.

So we need to divide 2 ^2

and 11 from the factorization to make 1188 a perfect cube.

Hence, the smallest number by which 1188 must be divided to obtain a perfect cube is 2^2 ×11 =44.

Hence the answer is 44

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