Question

If n is the smallest natural number such that n^2 +2n^2 +3n^2? +99n^2 is a perfect cube, then the number of digit in n is:​

Answer 1

Given:

n is the smallest natural number such that n^2 +2n^2 +3n^2? +99n^2 is a perfect cube

To Find:

The number of digit in n.

Solution:

We have the sum :  n^2 +2n^2 +3n^2 ....... +99n^2

• Sum = n² ( 1 + 2 + 3 + 4 + ..... + 99 )

   ( i )Sum of first n natural numbers  = n ( n + 1 ) /2

Therefore ,

• Sum = n² x  99 x 100 /2
• Sum = n²  x 9 x 11 x 25 x 2
• Sum = n² x 2 x 3² x 5² x 11 .

Now its given that the sum is a perfect cube .

• Hence if we take cube root of the sum , each of its factors should be an integer or a natural number.

Keeping this in mind,

  ( ii ) We have to make each factor 2, 3, 5 and 11 to power of 3 or multiples of 3.

• 3 and 5 are already squares.
• So 3 x 5 is required.
• 2 and 11 are raised to 1.
• So  we need 2²x11² for cubes of 2 and 11.

Also we have n² .

So n² = 3 x 5 x 2²x 11² , but this wont give square root of n² ,n as a natural number.

• Hence we need to check for higher powers of 2,3,5 and 11 which are multiples of 3 .

   

   ( iii )Let n = $2^{a} 3^{b} 5^{c} 11^{d}$  

Therefore ,

• n² = $2^{2a} 3^{2b} 5^{2c} 11^{2d}$

• Sum = $2^{2a} 3^{2b} 5^{2c} 11^{2d}$ x 2 x 3² x 5² x 11 .
• Sum  = $2^{2a+1}3^{2b + 2}5^{2c + 2}11^{2d + 1}$
• Least values a, b, c, and d such that

• 2a + 1 = multiple of 3 = 3

• 2b + 2 = multiple of 3 = 6

• 2c+ 2 = multiple of 3  = 6
• 2d + 1 = multiple of 3 = 3

Therefore ,

• a = 1
• b = 2
• c = 2
• d = 1

Therefore,

• n = 2 x 3² x 5² x 11 = 2 x 9 x 25 x 11 = 4950.

The number of digit in n is 4.