* Find a three digit number which, when reversed, becomes equal to 17 times the square of its cube root.
Answers
Answer: 216
Step-by-step explanation:
Let us consider the first, second & third digits of the three digit number be “x”, “y” & “z” respectively.
Then the 3-digit number will be 100x + 10y +z.
When reversed, the 3-digit number will become 100z + 10y + x.
According to the question, we have the equation as
100z + 10y + x = 17 * [100x + 10y +z]^(2/3) .... (i)
Now, in order to solve for the above equation,
1. Firstly, we will consider that the right side of the equation contains a cube root. So, we have to write down all 3-digit numbers having perfect cubes which are as follows:
5^3 = 125
6^3 = 216
7^3 = 343
8^3 = 512
9^3 = 729
10^3 = 1000 …… which is not a 3-digit number
3. Secondly, we will look if 17 times the square of the 3-digit number’s cube root gives the reversed order of the number.
17 * [(5^3)^2/3] = 425
17 * [(6^3)^2/3] = 612
17 * [(7^3)^2/3] = 833
17* [(8^3)^2/3] = 1088 …. Not a 3-digit number
3. Lets consider 216 as the 3 digit number and solve for the right side of the equation (i)
17 * [100x + 10y +z]^(2/3)
= 17 * [216]^(2/3)
= 17 * ∛(216^2)
= 17 * ∛46656
= 17 * 36
= 612
Therefore, from above we can see that only 216 satisfies the question given as its square of cube root is 612, which is the reverse order of 216.
Hence, the 3-digit number is 216.
Answer:
Step-by-step explanation: