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jacobwilliamroy

You can check the properties of digital roots on wikipedia:

https://en.wikipedia.org/wiki/Digital_root#Properties

Example from ASVAB practice math test:

(x+4)(x+4) =

A. x^2+16x+8

B. x^2+16x+16

C. x^2+8x+16

D. x^2+8x+8

Since we don't have to solve for X in this problem, we can just assume x is 1, which would make the digital root of the problem expression 7. Assuming x = 1, the digital roots of A, B, C and D are 7, 6, 7 and 8 respectively. C is the answer because its last term is the square of 4.

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Perhaps I am being dumb, but I don't see what the digital root brings to this.

You can do the same thing without the digital root. substitute 1 into all the possible answers gives: A: 25 B: 33 C: 25 D: 17

Substitute 1 into the question gives 25. Therefore only A and C are possible answers and it must be C because 4*4 = 16

This is the same process as your answer but without the digital root!

Maybe is case of large numbers you can compute digital root easier than the answer.

For example, 25 * 26 = 650, digital root 2. Or take digital root of 25 and 26 first, get 7 * 8 = 56, same digital root 2.

I suppose such is possible without digital roots. Let's try the same problem, but this time we'll multiply our constants by 3,571:

(x + 14284) (x + 14284)=

A: x^2 + 204032656x + 28568

B: x^2 + 204032656x + 204032656

C: x^2 + 28568x + 204032656

D: x^2 + 204032656x + 204032656

The digital root of the problem expression is 4. The digital roots of A, B, C and D are 4, 6, 4 and 3 respectively. The digital root of the square of 14284 is 1, and the digital roots of the last term of A and C are 2 and 1, respectively. The answer is C.

Of course, it is quicker just to multiply and add the last digit of the constant to find the only possible answer (or count digits to get an order of magnitude estimate, which will also show C as the only possible answer).

When a problem of this form uses big enough numbers to warrant a shortcut, digital root is almost always a suboptimal shortcut.

Yes you are the heuristics master!

Agreed. The real trick here is knowing that, since we don't need to find x, we can sub in an arbitrary x and do the comparison numerically instead of algebraically.

Of course there's a tiny chance that one of the other equations would be equal at your arbitrary x value. If so and you get two 'right' answers just try again with a different x.

It is the DeMillo-Lipton-Schwartz–Zippel lemma

If two polynomials evaluate to equal numbers with random variables values, the polynomials are almost certainly equal themselves

Ofc, x = 0 and x = 1 are hardly random

I can't decide if I've just seen magic, or conjuring. I'm gonna have to go away and think about this.

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