The sum of any three distinct natural numbers arranged in ascending order is 200 such that the second number is a perfect cube. How many possible values are there for this number ?

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3$
The second number is a perfect cube. It could be $1, 8, 27, 64, 125$
but, it cannot be $1$ as it is the second number
It cannot be $125$ as the third number would be greater than $125$ and then sum will become greater than $200$
Therefore, the second number can be $8, 27, 64$
Hence, option 'B' is correct.

Suggest Corrections

Similar questions

Q. Product of the distinct digits of a natural number is 60. How many such numbers are possible?

Q. How many natural numbers

n

are there such that

n! + 10

is a perfect square?

Q. N is a natural number, then how many values of N are possible such that

\frac{6 N^{3} + 3 N^{2} + N + 24}{N}

is also a Natural Number?

Q. How many such digits are there in the number 5261983 each of which is as far away from the beginning of the number as then the digits are arranged in ascending order within the number ?

How many such pairs of digits are there in the number 132546798 each of which has as many digits between them in the number as when they are arranged in ascending order?

Join BYJU'S Learning Program