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What is the missing number in the following set?

$$3, 4, ..., 24, 43, 68$$

-- Sawyer, "Mathematician's Delight"

The author states that it must be a number between 5 and 23, and that the worst case is to calculate the difference tables $$\Delta_\Upsilon,\Delta^2_\Upsilon$$ etc. for all the numbers between 5 and 23 (the right number will be the one that results in all zeros at some difference row). Any recommendations to find a shorter solution?

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up vote 3 down vote accepted

The number is $11$. Observe that $4-3=1$, $11-4=7$, $24-11=13$, $43-24=19$ and $68-43=25$.

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Hey hey! 12 K!! – amWhy Dec 24 '12 at 0:30

For the difference table to have an all-zero row, the numbers must be values of some polynomial $p(x)$, and that means you can use Lagrange interpolation. I'll set it up this way: $$\matrix{x&-2&-1&0&1&2&3\cr p(x)&3&4&?&24&43&68\cr}$$

Now, consider the polynomial $A(x)=(x+1)(x-1)(x-2)(x-3)$. We get $A(-2)=-60$, $A(x)=0$ for all the other values of $x$. So if I let $A^*(x)=-(1/20)A(x)$, then $A^*(-2)=p(-2)=3$, and still $A^*(x)=0$ for $x=-1,1,2,3$.

Now use the same ideas to make a polynomial $B^*(x)$ with $B^*(-1)=p(-1)=4$ and $B^*(x)=0$ for $x=-2,1,2,3$, and then $C^*(x),D^*(x),E^*(x)$ to get the right values at $x=1,2,3$, respectively. Then if you add these five polynomials, you get a single polynomial $P(x)$ that agrees with $p(x)$ at $x=-2,-1,1,2,3$. Then the number you want is $P(0)$.

As it happens, that's overkill for this problem, as the answer turns out to be a quadratic polynomial. But if you don't know that, and you're not clever enough to see it the way Jasper did in the other answer, the Lagrange method has the advantage of being completely mechanical and always giving an answer.

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