This is why I love the number 7. It’s the first real prime number. All the others are “first”…1?2?3?5? No, those aren’t prime numbers, they’re “first” in a long line of not-prime numbers.
Then you get to 7. Is 27943 divisible by 7? If you take away 3 is it? If you add 4 is?
I have no clue, give me 10 minutes or a calculator is the only answer
That’s what a real prime number is.
Take the last digit of the number, double it and subtract it from the rest. If that new number is divisible by 7, the original one is as well. For your example:
2794 - 6 = 2788
I know 2800 is divisible by seven, so 2788 is not. Thus 27943 is not divisible by 7.
Quick maff shows that neither subtracting 3 or adding 4 will make the original number divisible by 7. Adding 1 or subtracting 6 will tho.
Quick check for divisibility: subtract 7 from it. If the new number is divisible by 7, then the original number is too
Also, any number whose digits sum to a multiple of 3 is divisible by 3. For 51, 5+1=6, and 6 is a multiple of 3, so 51 can be cleanly divided by 3.
Does this also work the other way round, i.e. do all multiples of three have digits that sum to a multiple of 3? All the ones I’ve checked so far do, but is it proven?
Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + … + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you’re left with n = a_0 + a_1 + … + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + … + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3
I’d forgotten this trick. It works for large numbers too.
122,300,223÷3 = 40,766, 741
1+2+2+3+2+2+3 = 15
threw up and died while reading this
I wish I could read 😞
Just squint and wing it.
Any number where the individual digits add up to a number divisible by ‘3’ is divisible by 3.
51 = 5+1 = 6, which is divisible by three.
Try it, you’ll see it always works.
There are tricks like that for a lot of numbers. For 7, chop off the last digit, double it and add it to what’s left. Repeat as required. If the result is divisible by 7 then the original number was. eg: 356 -> 35+12=47 not db7. 357 =>35+14 both db7 so we don’t even need to do the add.
