Let the three numbers in GP: a/r, a, ar---------(A)
Where '^' is power of . . .
Sum of these numbers are:
a/r +a +ar = 38
a(1\r+1+r) = 38 ---- (1)
Product of these numbers are:
a^3 = 1728
= (12)^3
a = 12
Putting the value of a in (1) you will get:
12(1\r+1+r) = 38
And factorising, we get
r = 2/3 or r = 3/2
Sub. the r and a value in (A), we get
8,12,18 or 18,12,8. When a = 12
And smallest no. is 8.
(12)144 = 1728
1728 is a pure number. Without units it has no meaning in volumetric measurements.
To determine if 1728 is divisible by 6, we need to check if the sum of its digits is divisible by 3 and if it is an even number. The sum of the digits of 1728 is 1+7+2+8=18, which is divisible by 3. Additionally, the last digit of 1728 is 8, which is an even number. Therefore, 1728 is divisible by 6.
It is: 12*12*12 = 1728
24√3 or 41.5692194 or 41.57, to the justified number of significant digits since 1728 has only four.
the cube root of 1728 is 12
As a product of its prime factors in exponents: 2^6 times 3^3 = 1728
(12)144 = 1728
The number 12.
Ramanujan number is the smallest natural number that can be can be expressed as a sum of two perfect cubes in two different ways:- 123 + 13 = 1728+1 =1729 103 + 93 = 1000+729 =1729
1728 is even so it cannot have an odd cube root.
1728 is a pure number. Without units it has no meaning in volumetric measurements.
1728/60=28 and left 48number is written as 28,48
To determine if 1728 is divisible by 6, we need to check if the sum of its digits is divisible by 3 and if it is an even number. The sum of the digits of 1728 is 1+7+2+8=18, which is divisible by 3. Additionally, the last digit of 1728 is 8, which is an even number. Therefore, 1728 is divisible by 6.
It is: 12*12*12 = 1728
3√1728 = 12
24√3 or 41.5692194 or 41.57, to the justified number of significant digits since 1728 has only four.