Question

70304 is divided by the smallest number so that the result is a perfect cube. Find the cube root of the resulting number.The sum of four consecutive odd natural numbers is 136. Find the largest of these numbers.

Answer 1

Answer:

70304/32 =2197

2197=13power3

Answer 2

We start with finding the prime factors of the given number.

$70304 = 2^{5}*13^{3}$

$70304 = 4*2^{3}*13^{3} \\70304 = 4*26^{3}$

We can clearly see that the number is the product of 4 and the cube of 26.

Therefore, when the given number is divided by 4, we get a perfect cube of 26 i.e. 17576.

Let the first odd natural number be $2x + 1$.

The second odd number will be $2x + 3$ .

The four consecutive numbers will be $2x + 1, 2x + 3, 2x + 5, 2x + 7.$

It is given that the sum of these numbers is 136. Then,

            $2x + 1 + 2x + 3 + 2x + 5 + 2x + 7 = 136\\8x + 16 = 136\\8x = 120\\x = 15$

The first odd number in the sequence is $2x + 1 = 2*15 + 1 = 31$

The complete sequence of numbers > 31,33,35,37.

Therefore, the largest number in the sequence is 37.