2
$\begingroup$

https://en.m.wikipedia.org/wiki/Carry_(arithmetic)

Definition of carry in Wikipedia

I'm guessing the answer is no but just to be sure I wanted to know of there is a way to predict the amount of carry overs one can get by doing a sum in base 10.

Given a natural number $a$, for example, can I predict the amount of carry overs I will get by doing $ a^2 + 2a +1$? I want to know if it is possible to do this without actually computing the sum.

Thanks in advance

Edit: To clarify, the number of carries in a sum indicates the amount of times one position adds 10 or more. I'm finding this hard to explain so here goes an example:

When you were at school you were probably taught to sum by adding in each position, and if the sum of the numbers in that position exceeds 10, you "carry" a 1 to the next position to the left, leaving in the original position the rightmost digit of your sum.

So for example 5+6 has 1 carry since you add two numbers in the $10^0$ position and carry a 1 to the $10^1$ position

$\endgroup$
  • $\begingroup$ Can you define a "carry"? $\endgroup$ – Austin Weaver Oct 31 '17 at 14:52
  • 2
    $\begingroup$ Would that count have to include the number of carries for computing $2a$ or $a^2$? $\endgroup$ – cronos2 Oct 31 '17 at 14:52
  • $\begingroup$ Deleted my previous comment. No, do not include those carries. Just the ones from doing the original sum. I'm trying to check whether computing those carries would change something $\endgroup$ – Francisco José Letterio Oct 31 '17 at 15:13
  • 1
    $\begingroup$ So you're looking for corresponding digits in $a^2$ and $2a$ either where the sum of the two digits is greater than $9$ or where the sum is exactly $9$ and there is a carry from the next place to the right. My guess is that there is not a way to do this that is significantly more efficient in the general case than actually doing the sum. (Possibly in actual computing there is a way to gain a few processing cycles by ignoring some of the results of the addition and looking only at the carry.) $\endgroup$ – David K Oct 31 '17 at 15:16
  • $\begingroup$ Yeah I was guessing that actually predicting this for any number a would be too good to be true. I guess I could always try to do some calculations heuristically with a pc. $\endgroup$ – Francisco José Letterio Oct 31 '17 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.