tdn writes:
> Any quick trick to knowing the answer without working it all out step by step?
Buy a calculator, since that’s easier than learning a trick.
Of course, if you attend to the next level of abstraction, cmyk’s finger trick works for any base -1 (add or subtract fingers accordingly to total 10 in whatever base you’re using).
That reminds me of one of my favorite math tricks.
To determine the sum of a series of numbers from 1 through x:
Divide x by 2
Add 0.5
Multiply by x
For example, for the range 1 through 7:
7 / 2 = 3.5
3.5 + 0.5 = 4
4 x 7 = 28
so 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
Other neat number tricks:
A three-digit number with all the same digits is divisible by 37.
A four digit-number in the form of xxyy or xyyx is divisble by 11.
A four-digit number where the first two digits are double the last digits (as in 6834) is divisible by 67. Same for three digit-numbers where the first digit is double the last two, like 804.
Multiply 142857 by 2,3,4,5 and 6, and the products will have the same digits in different orders. Multiply it by 7, and you get 999999.
Wait… 1+2+3 = 6 (Six is an even number.)
123 / 3 = 41 Check
123 / 6 = 20 1/2 
I hate it when threads about third grade math confuse me.
Obviously not, as 123 is a counterexample. More generally, we have that any number whose digits sum to a multiple of 3 is divisible by 3. Since addition is commutative, any permutation of those digits is also divisible by 3. But the number is only divisible by 2 if the last digit is divisible by 2, and that does not hold under an arbitrary permutation.
One other trick that may come in handy: a number is divisible by 2[sup]n[/sup] exactly when the last n digits are divisible by 2[sup]n[/sup]. I’ll leave it as an exercise to the reader to show why this is true.
If the number itself is also even (i.e., if its last digit is also even), then it’s divisible by 6.
If a number is divisible by 2 and 3 then it is also divisible by 6. I think that’s easier to make clear.
And, if the last 2 digits are divisible by 4, then the entire number is divisible by 4.
If the last 3 digits are divisible by 8, then the entire number is divisible by 8.
To multiply by 5, take half and then move the decimal point once to the right (add a zero if it’s a whole number result). Ex: 24 x 5 = 120 as 12 is half of 24.
To multiply by 11, add the digits together to get a digit that will go inbetween. Ex:
25 x 11 = 275 Keep the 2 and 5 at the start and end of the number respectively and add them to get the 7 that comes between. If the result is more than 9, then you simply carry the one onto the hundreds digit. You can do this with as many digits as you wish.
Soooo, to multiply by 55 (I’m sure you do this all the time) you can combine those two tricks. 34 x 55 = 1870. Half of 34 is 17. Add them together to get the 8 in the middle 187. Add the zero from the “5 trick”.
Squaring a number that ends in 5 is easy as well. 35^2 = 1225. The answer always ends in 25. To get the first part, take the non-five digit and multiply it by the whole number that is one greater than what you have. 3 x4 = 12, then add on the 25 to get 1225. So, 65^2 would be 4225.
Incidentally, the trick generated for 6 by the uniform method of generating these tricks is to subtract all the digits * 2, except the rightmost one which you just add; the result will have the same remainder when divided by 6 as the original number. Thus, 13598157 comes out to -2*(1+3+5+9+8+1+5) + 7 = -57, which in turn comes out to -(-2*5 + 7) = -(-3) = 3, which is the original number’s remainder when dividing by 6.
Of course, if it’s easier for you, -2 above can be replaced by 4. But then again, the easiest thing is checking divisibility by 2 and by 3 separately. Just mentioning this for completeness’s sake.
Speaking of finger tricks, I once heard there is a way to remember which months have 30 or 31 days by counting on your fingers. Anyone know this trick? I mean, I know the “30 days hath September…” rhyme, but I’m curious about the finger method.
Yeah, curl a hand into a fist and look at your lowest knuckles (not counting the thumb): bump, valley, bump, valley, bump, valley, bump. Go back to the beginning after you get to the end, and think of bumps as 31 days and valleys as < 31 days (i.e., 30 except for February’s 28/29), and this matches the sequence of day-lengths for the calendar from January through December.
Technically, it’s a knuckle trick.
Start with the knuckle where your index finger meets your hand (the metacarpal joint?), call that January. Next you have the valley between your first knuckle and your second, that’s February. Knuckle — January; valley — February; knuckle — March etc.
When you run out of knuckles, start over with the index finger of your other hand. (You can of course use the same hand if you want.)
Months that land on a knuckle have 31; those that land in a valley have 30 except for February.
I was taught the “divisible by three” trick but figured out the “multiplying by nine” thing on my own. Always thought it was pretty cool.
Simple explanations:
Yes, because 111 is divisible by 37 and the others are clearly all multiples of this.
Yes, because 1100, 0011, 1001, and 0110 are all divisible by 11 and the others are clearly all (integer) linear combinations of these.
Yes, because 0201 is divisible by 67, and the others are clearly all multiples of this.
Yes, this is just the special case of the above where the first of the four digits is 0.
Also, I linked to it before, but I might as well give a proof/explanation of how the divisibility tricks work. There are a couple ways to see it, as often is the case, but here’s one that’s in the style of “high school/college algebra”. I’ll first demonstrate for the case of 9 specifically, though the ideas generalize straightforwardly:
Consider an arbitrary polynomial Q(x) = A + Bx + Cx^2 + Dx^3 + … . Taking advantage of the fact that A = Q(0), we can rewrite this as Q(x) = Q(0) + x*(B + Cx + Dx^2 + …). Since Q(x) and Q(0) differ by a multiple of x, we see that Q(x) is divisible by x if and only if Q(0) is [and, indeed, the two will always have the same remainder even otherwise].
Now, take a number in decimal representation [e.g., 2356]. This is essentially short-hand for some polynomial P(x) evaluated at 10 [with our example, let P(x) = 2x^3 + 3x^2 + 5x + 6; the number we wrote is then P(10)]. Define the polynomial Q(x) as P(x+1), so that our original number is Q(9). Now, using the result from above, we see that 9 divides Q(9) [our original number] if and only if 9 divides Q(0) [= P(1) = the sum of the digits of our original number], establishing our desired result.
More generally, for any divisor, the tricks all amount to simply going through the expanded decimal representation of our number [e.g., 210^3 + 310^2 + 510+6] and replacing all those powers of 10 by swapping 10 for something which differs from it by a multiple of our divisor. Thus, we can replace it with 1 for divisors of 9 and 3, we can replace it with 0 for divisors of 2, 5, and 10, we can replace it with 2 for a divisor of 4 (and note that from 10^2 on, we can replace it with 0), with 3 for a divisor of 7 (and note that actually 1, 10, 10^3, …, can be replaced with 1, 3, 2, …), with -2 or 4 for a divisor of 6, etc. All the divisibility tricks in this thread so far are of basically this form (that or noting that divisibility by ab is the same as divisibility of a conjoined with divisibility by b, when a and b share no prime factors).
For 11, I’ve always heard that it was just move the number over and then add. Carry over as needed. (I think it’s the same as the other 11 tricks here, but just explained differently)
So 377 * 11 gives us
00377
+3770
4147
All you’re really doing when you multiply by 11 is multiplying by 10 and then adding the original number back to it.
One “trick” that I actually use on a regular basis… for 15% if you’re tipping at a restaurant just take
10% -> move the decimal over to the left
- 5% -> half of the above
Similarly, 20% is 2 * 10%, which you get the same way as above.
Lest anyone suffer in suspense …
Disregarding the last n digits, since we already know they divide by 2[sup]n[/sup], we’re left with a number of the form r*10[sup]n[/sup], for some r. 10[sup]n[/sup]=5[sup]n[/sup]2[sup]n[/sup]; 10[sup]n[/sup] is divisible exactly by 2[sup]n[/sup], and therefore so is r10[sup]n[/sup].