I never would have remembered this bit of the multiplication table were it not for this:

The product of multiplying by nine, when all the digits are added together, will always equal nine. To be honest, I don’t know how far this rule applies* but, for the purposes of third grade multiplication, it always worked. An added trick to remember things was that, when multiplying numbers two through nine by nine, the product would not only equal nine when added up but the first digit would always be one less than whatever it is I was multiplying.

Examples: 9 x 2 = (one less than 2 is 1+8=9) 18, 9 x 6 = 54, etc.

Also, once you get to five, the answers are all mirror images (18, 27, 36, 45, 54, 63, 72, 81).

I do not believe for one second that I came up with this all on my own when I was 8 years old, but nobody I know has ever realized that this is how nine works including people who grew up with me and went to the same school I did.

Ever heard of it?

*The biggest random number I multiplied by nine in google that had digits that added up to nine was 2678 (24102). The ones that didn’t add up to nine added up to multiples of nine, though. Interesting.

Yep, I was taught this too. Similarly, any number who’s digits ultimately add up to a number divisible by three, is also divisible by three. So 123 → 1+2+3 = 6, so 123 is evenly divisible by three.

Yes, and you can come up with rules like these for any number (even 7, though people will tell you otherwise). As the OP notes at the end, they won’t always add up to exactly 9 right away; rather, they will add up to a multiple of 9, so that, repeatedly carrying out the process till you come down to only one digit, you get 9 (or 0, if you started with 0…). And, of course, it’s important to note that this goes both ways: if you get 9 by doing this, then your original number was a multiple of 9 as well.

Why this works is a simple exercise in algebra which I’ll link to if I can find where we’ve discussed this before. Ah, here it is.

And here’s the trick for 7, though not quite as neat:
Split your decimal representation into digit-triplets starting from the right (thus, 23256881632 would split into 023 256 881 632). Now, alternately add and subtract these; the result will be divisible by 7 if and only if the original number was. (So keep carrying this out till you get just one triplet, and just check if that one is divisible by 7)

The digits of Multiples of 3s always add up to 3 6 or 9
Multiples of 5 of course always end in 0 or 5
Multiples of 2 are always even.
The digits of Multiples of 9 always add up to 9 if you keep adding until down to a single digit.

For powers of 2 (Binary) for bytes
2[sup]10[/sup] is a kilobyte
2[sup]20[/sup] is a megabyte
2[sup]30[/sup] is a gigabyte
etc.

I never knew the 7’s trick. I am not sure how useful it is though.

Along the lines of what cmyk posted, 9*x for x<10 is always x-(10x). Algebraically, this is tautologous, but it’s a useful trick to teach folks who are otherwise struggling with their 9s.

A useful trick with 11s is that to get 11*x, you simply add the digits of x (for example, 13) and put them in between x’s digits:

11*13=1(1+3)3=143.

This gets more complicated when you have an x with more than two digits or whose digits add up to more than 9:

11*19=1(1+9)9=209 because 1+9=10. Imagine that the hole between 1 and 9 is a zero, and add the 1 from 19 and the 1 from 10 (1+9). You then get 209.

But what about 11*199? 2189, because:

1+9=10, so 2 (1 from 19 and 1 from 10)

18 from 9+9 (the second and third digits of 199), as far as I can tell

Well, one can express the same trick in so many ways, of course. I buried it for simplicity, but part of the trick is also that a triplet ABC has the same remainder when divided by 7 as 2A + 3B + C.

Thus, a more detailed version of the same trick goes like this: moving from right to left, do “Add once, add three times, add twice, subtract once, subtract three times, subtract twice, repeat”. And at any point along the way, one can simplify numbers to their remainder mod 7 as one likes.

For example, let’s take your post’s ID number: 10181595. We can think of 9s as 2s and 8s as 1s; thus this turns into 10111525. Moving from right to left, we have 5 + 32 (= 11 = 4) + 25 (= 14 = 0) - 1 (= -1 = 6) - 31 (= 3) - 21 (= 1) + 0 (= 1) + 3*1 (= 4). Thus, 10181595’s remainder when divided by 7 is 4.

This will at least mitigate worries about three digit additions and subtractions and carrying and so forth, by keeping everything small.

Ideally. In reality, though, disks are sold with capacities closer to the base-10 kilo-, mega-, and giga-, which are progressively smaller than their base-2 counterparts as you move up the scale (1,024 - 1,000 = 24, 1,048,576 - 1,000,000 = 48,576, 1,073,741,824 - 1,000,000,000 = 73,741,824, and so on). Plus, of course, there are outright lies such as a ‘megabyte’ defined as 1,024,000 bytes (or 1,000 base-2 kilobytes) (This, incidentally, is where we get the 1.44 MB capacity for 3.5"* stiff floppies).

*(OK, I’m not going into that. Metrication can bite my shiny metal ass.)

I was taught the same thing! And now I have that Square One song stuck in my head, too. Man, I was a dorky kid, but I loved that show with a peculiar, mathy passion.

If I have a dollar, then double it, I have 2 dollars. If I double it again, I have 4 dollars. Doubled twice more, I have 16. If I double it 17 more times I have…

Any quick trick to knowing the answer without working it all out step by step?

Well the 10th doubling is a little over a million. That is the same as 2[sup]10[/sup]
So as a quick rule of thumb every 10 doublings would add 3 zeros to the number.

You have 2[sup]4[/sup], you’re multiplying by 2[sup]17[/sup], so the total is going to be 2[sup]21[/sup]. The general rule is that x[sup]a[/sup] * x[sup]b[/sup] = x[sup]a + b[/sup].