Home Puzzles arithmetic – Powerful quantity methods

### arithmetic – Powerful quantity methods

Imagine I’ve two nonnegative integers, $$a$$ and $$b$$. Can you discover the final digit of the sum $$a+b$$ given solely the final digits of $$a$$ and $$b$$? More typically, can you discover the final $$n$$ digits of $$a+b$$ given solely the final $$n$$ digits of $$a$$ and $$b$$? The reply to each questions is sure: for instance, any variety of the shape $$…123$$ added to any variety of the shape $$…789$$ all the time leads to plenty of the shape $$…912$$.

What in regards to the final $$n$$ digits of the product $$atimes b$$? There can also be a singular reply on this case: for instance, $$(…47) occasions (…38) = …86$$ all the time.

What in regards to the final $$n$$ digits of the ability $$a^b$$ (the place we take $$0^0=1$$)? Sadly, right here the entire thing breaks down: besides in very uncommon circumstances, the final digits of an influence can not basically be decided from the final digits of the bottom and exponent. For instance, $$13^{14} = …9$$, however $$13^{24}=…1$$, so $$(…3)^{…4}$$ isn’t uniquely decided. So our bizarre decimal notation behaves effectively with respect to sums and merchandise, however not with respect to exponentiation. That’s a disgrace! Could this “subject” maybe be fastened by altering our quantity system to one thing extra… highly effective?

Let $$operatorname{final}_n a$$ denote the quantity obtained by taking the final $$n$$ digits of the quantity $$a$$ (and $$operatorname{final}_n a = a$$ if $$a$$ has lower than $$n$$ digits). For instance, we have now $$operatorname{final}_3 14835 = 835$$, $$operatorname{final}_4 57 = 57$$ and $$operatorname{final}_2 302 = 02 = 2$$. We’ve seen that $$operatorname{final}_n (a+b) = operatorname{final}_n(operatorname{final}_n a + operatorname{final}_n b)$$ and $$operatorname{final}_n (atimes b) = operatorname{final}_n(operatorname{final}_n a occasions operatorname{final}_n b)$$ for all $$a, b$$ and all $$n>0$$. We’ll say {that a} quantity system is a highly effective quantity system if moreover we have now $$operatorname{final}_n (a^b) = operatorname{final}_n((operatorname{final}_n a)^{operatorname{final}_n b}).$$

Unfortunately, bizarre base-$$B$$ positional notation is not highly effective for any $$B$$ (are you able to see why?), so we’re compelled to show to extra unique methods:

• The bijective base-$$B$$ quantity methods are like bizarre base-$$B$$ numbers, however as a substitute of the digits going from $$0$$ to $$B-1$$, they go from $$1$$ to $$B$$ (and nil is represented by the empty string). The most well-known occasion is unary notation or bijective base-$$1$$, during which a quantity is represented by writing that many $$1$$s in succession. For instance, the checklist of numbers in bijective base-$$10$$ notation goes like $$textual content{(empty)}, 1, 2, 3, 4, 5, 6, 7, 8, 9, mathrm A, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1{mathrm A}, 21, ldots$$ (the place the digit $$mathrm A$$ represents ten).

• The factorial quantity system is a system which isn’t tied to any base particularly; as a substitute, it’s such that the $$okay$$th digit is allowed to differ from $$0$$ to $$k-1$$ (so in idea this base wants infinitely many symbols, however solely a finite quantity is required to signify a given quantity). The checklist of optimistic numbers on this notation goes like $$0, 10, 100, 110, 200, 210, 1000, 1010, 1100, 1110, 1200, 1210, 2000, 2010, 2100, 2110, 2200, 2210, 3000, 3010, ldots$$. This quantity system is named as such as a result of the factorial $$N! = N occasions (N-1) occasions cdots occasions 2 occasions 1$$ of any quantity $$N$$ is definitely expressed on this notation as an $$1$$ adopted by $$N$$ zeros. For instance, $$6!$$ is expressed as $$1000000$$.

• Finally, we are able to mix the concepts of the bijective and factorial quantity methods by defining the bijectorial quantity system: a system such that the $$okay$$th digit is allowed to differ from $$1$$ to $$okay$$. The numbers on this notation go like $$textual content{(empty)}, 1, 11, 21, 111, 121, 211, 221, 311, 321, 1111, ldots$$

Can you discover all highly effective quantity methods among the many talked about ones (bijective base-$$B$$ for some $$B$$, factorial, bijectorial)?

Hint:

There are finitely many.