X=4a+3b, If a is a two digit cube number and b is a two digit square number then what is the lowest possible value for X?

In order to find the lowest value for X we first need to find the lowest possible value for a and b.

Let us start off with a:

If a is the lowest possible cube number we can start by checking the first couple of cube numbers and we get:

- 1^3 = 1

  • 2^3 = 8
  • 3^3 = 27

Hence, the lowest possible two digit cube number we can get is 27, so a=27. Now let us check for b, what is the lowest possible two digit square number:

- 1^2 = 1

  • 2^2 = 4
  • 3^2 = 9
  • 4^2 = 16

Hence, the lowest possible square number we can get is 16, so b=16.

Now the math is easy, we just plug our values for a and b into the equation to find the answer:

X = 427 + 316 = 108 + 48 = 156

So the answer is 16.
 

SI
Answered by Silje I. Maths tutor

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