Perfect What?

Number Theory Level 3

Given $a,b,c,d,e$ are consecutive positive integers such that their sum is a perfect cube, and the sum of the middle three terms is a perfect square.

What is the value of $3c - 10$ ?

The answer is 2015.

1 solution

Sravan Chinta
Dec 15, 2014

$Given\quad that\quad a,b,c,d,e\quad are\quad consecutive\quad integers\\ let\quad a=(x-2),b=(x-1),c=x,d=(x+1),e=(x+2)\\ \Rightarrow a+b+c+d+e=5x\longrightarrow (1)\\ \quad \quad b+c+d=3x\quad \quad \quad \quad \quad \longrightarrow (2)\\ \Rightarrow let\quad 5x={ p }^{ 3 }\quad \& \quad 3x={ q }^{ 2 }\\ \Rightarrow minimum\quad value\quad of\quad x\quad =27\times 25=675=c\\ \therefore \quad minimum\quad value\quad of\quad 3c-10=2015$

why can't we taken x=0 (all the conditions are satisfied too) , in my first attempt i tried this. @Sravan Chinta can you help me

U Z - 6 years, 6 months ago

I agree with you, Megh. I also obtained x=0 as the solution.

Let {a,b,c,d,e} = {x-2,x-1,x,x+1,x+2} with x an integer.

Since c==x, we must minimize x to minimize, 3x-10.

Then 3x must be a perfect square and 5x is a perfect cube.

Since 3x >= 0, 3x and 5x are the minimal perfect squares and cubes.

So, the minimum value of 3c-10 is $\boxed{ -10}$ .

Michael Fischer - 6 years, 5 months ago

Thanks. I have updated the phrasing of the problem.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “dot dot dot” menu in the lower right corner. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 6 years, 2 months ago

I think that the definition for a perfect square is a number that can be obtained by multiplying a natural number to itself. So 0 doesn't fit

Archit Boobna - 6 years, 5 months ago

0 is a perfect square, since $0 = 0^2$ .

Calvin Lin Staff - 6 years, 2 months ago

Perfect What?

The answer is 2015.

1 solution

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