Maths Olympiad Prep

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Problem 838

AMC 12 late, AIME early
Algebra Difficulty 4.5 Find the answer Harvard-MIT November Tournament · United States

w,x,y,zw, x, y, z are real numbers such that
w+x+y+z=52w+4x+8y+16z=73w+9x+27y+81z=114w+16x+64y+256z=1 \begin{aligned} w+x+y+z & =5 \\ 2 w+4 x+8 y+16 z & =7 \\ 3 w+9 x+27 y+81 z & =11 \\ 4 w+16 x+64 y+256 z & =1 \end{aligned}
What is the value of 5w+25x+125y+625z?5 w+25 x+125 y+625 z ?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Solution:

Answer: 60-60

We note this system of equations is equivalent to evaluating the polynomial (in aa) P(a)=wa+xa2+ya3+za4P(a) = w a + x a^{2} + y a^{3} + z a^{4} at 1,2,31, 2, 3, and 44. We know that P(0)=0P(0) = 0, P(1)=5P(1) = 5, P(2)=7P(2) = 7, P(3)=11P(3) = 11, P(4)=1P(4) = 1.

The finite difference of a polynomial ff is f(n+1)f(n)f(n+1) - f(n), which is a polynomial with degree one less than the degree of ff. The second, third, etc. finite differences come from applying this operation repeatedly. The fourth finite difference of this polynomial is constant because this is a fourth degree polynomial.

Repeatedly applying finite differences, we get

Figure 1

and we see that the fourth finite difference is 21-21. We can extend this table, knowing that the fourth finite difference is always 21-21, and we find that P(5)=60P(5) = -60.

The complete table is

057111-60
524-10-61
-32-14-51
5-16-37
-21-21

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.