Welcome to my blog! I’m rather new to blogging, so please excuse any weird newbie traits I may have. My main reason for creating this blog is because I was going to respond to a comment on my status updates with a proof for a 42 fact I posted earlier, but the proof rapidly became far, far, far too long for a status update, so my solution was to create a blog and put the proof here instead. I’ll probably continue to post more facts about 42 here as well, and whatever else happens to interest me. So for the 0.1% of you out there who also love maths and the number 42, this is the place to be!

Anyway, the proof is for 420 being the smallest positive integer divisible by 1, 2, 3, 4, 5, 6 and 7. ‘Smallest’ is a key word here, as the proof will also show there are no positive integers smaller than 420 that are divisible by 1, 2, 3, 4, 5, 6 and 7. Anyway, all you really need to know for the proof are the divisibility tests for each of those numbers. I’ll quickly list them here:

1: Everything is divisible by 1, silly!

2: The number is even. More specifically, the units digit of the number is either a 2, 4, 6, 8 or 0.

3: The sum of the digits is divisible by 3 (e.g. to test 522, you’d add 5 + 2 + 2, which is 9. 9 is divisible by 3, so 522 must be divisible by 3)

4: The number created by the last two digits is divisible by 4 (e.g. to test 3644, you’d only look at the number made by the last two digits, 44. 44 is divisible by 4, so 3644 must be divisible by 4)

5: The units digit is either a 5 or 0.

6: The number satisfies the divisibility tests for both 2 and 3.

7: Take the units digit of the number and double it. Then, divide the original number by 10, rounding down to the nearest integer. Subtract the value you obtained by doubling the units digit from this new number. If the result is 0 or is divisible by 7, the original number is divisible by 7. (This explanation is rather long since I’m trying to avoid ambiguity, an example will probably really help in understanding it. To test 959, first double the units digit, 9, to get 18. Then, divide 959 by 10, rounding down, to get 95 (You always round down, so it’s not 96). Subtract 18 from 95 to get 77. 77 is divisible by 7, so 959 must be divisible by 7.)

Anyway, proof time:

First of all, note that it must be divisible by both 2 and 5. If the number is divisible by 2, it must be even, meaning its units digit is a 2, 4, 6, 8 or 0. If a number is divisible by 5, its units digit must be a 5 or a 0. Therefore, for it to be divisible by both 2 and 5, its units digit must be a 0, since that’s the only overlap between the possible units digit. Since the units digit must be 0, we have managed to restrict the possible options to multiples of 10.

Now, consider divisibility by 4. Therefore, the last two digits of the number must also be divisible by 4. However, since the units digit is forced to be 0, the only possible values for the last two digits that’s still divisible by 4 are 00, 20, 40, 60 and 80. Now we’ve restricted the possible options to multiples of 20.

Now consider divisibility by 3. Since the test involves adding the digits together, we can ignore the units digit in these tests since it is always 0. Doing so makes determining which numbers are divisible by 3 a simple matter of knowing your threes times table. For instance, if we wanted to test the first 6 numbers we’ve narrowed down our options to (20, 40, 60, 80, 100, 120), we just need to check if the number, after ignoring the units digit, is still divisible by 3. In this case, it means checking if 2, 4, 6, 8, 10 and 12 are divisible by 3. 6 and 12 are, but the rest aren’t, so 60 and 120 are divisible by 3 while 20, 40, 80 and 100 are not. In fact, it turns out that with the multiples of 20 we’ve narrowed down the choices to, every third number is divisible by 3, which results in narrowing down the options to multiples of 60.

At this point, the only thing left to check is divisibility by 7 (We can ignore 1 because everything is divisible by 1, silly! We can also ignore 6, because we’ve narrowed down our choices to numbers that are both multiples of 2 and 3, so they must all be multiples of 6 as well). We know the units digit of the remaining numbers is 0, which simplifies the test a lot; Multiplying 0 by 2 and subtracting the result, 0, from another number does nothing to that number, so we can ignore that step of the 7 divisibility test. All that’s left to do in the test is divide the numbers by 10 and see which one is divisible by 7. Since we’ve narrowed down our possible options to just multiples of 60, it is rather easy at this point to just check every value we still have left.

60/10=6, not divisible by 7.

120/10=12, not divisible by 7.

180/10=18, not divisible by 7.

240/10=24, not divisible by 7.

300/10=30, not divisible by 7.

360/10=36, not divisible by 7.

420/10=42, which is divisible by 7.

Therefore, 420 is the smallest positive integer divisible by 1, 2, 3, 4, 5, 6 and 7.

I love maths.