Given that 198 = 2 × 32 ×11; find the smallest integers ‘k’ such that 198k is a perfect square.?

Answer 1

198 = 2 x 3² x 11

k = 22

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If a number is a perfect square, then each of its prime factors has an even power (i.e. 2 or a multiple of 2).

198 = 2 x 3² x 11

if we times this by 2 x 11, we get

2² x 3² x 11² = (2 x 3 x 11)²

so k = 2 x 11 = 22

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Hope this helps!

Answer 2

Well, 198 is NOT 2 * 32 * 11. 198 = 2 * 3^2 * 11 I think that's what you mean.

Since the factor 3 has exponent 2 in the factorization of 198 into prime numbers, we must determine k so that the factorization of 198k containsll it's prime numbers with even positive exponents. Since it must contain the prime factors 2 and 11, the solution is k = 2 * 11 = 22.

Then, 198k = 2^2 * 3^2 * 11^2 = (2 * 3 *11)^2 = 66^2, a perfect square.

Answer 3

Divide 198 into prime factor, we have:
198 = 2 x 2 x 2 x 2 x 2 x 2 X 11 = 2^6 X 11.
Hence the smallest is 11

Answer 4

198=2*2*2*2*2*2*11
=2^2*2^2*2^2*11
therefore the the number should be multiplied by `11 to make it a perfect square

Answer 5

198 = 2*3^2*11

to have a perfect square you must multiply by 22 =2*11

so you have 2^2*3^2*11^2=4356

Answer 6

11
& ur =n is wrong