Find a three digit number which, when reversed, becomes equal to 17 times the square
of its cube root.
Answers
Answer:
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
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