Question

Find a six-digit multiple of 64 that consists only of the digits 1 and 2.

Answer 1

If it helps any 100032 is the smallest 6 digit multiple of 64

Answer 2

Oh. and 111104 is the smallest multiple where there are no more zeroes in the first 4 digits.

Answer 3

Just sharing information. The answer is going to be between \(64\cdot 1563\) and \(64\cdot 4687\). After that you get into the 300 thousands, which wont satisfy the conditions.

Answer 4

The ones digit must also be a 3 or an 8.

Answer 5

But, it has to consist of only 1 and 2

Answer 6

Er, the ones digit of the number that you multiply with 64.

Answer 7

i guess if you keep adding 320 to 100032 you\ll eventually get there

Answer 8

That's what I meantm my bad >.<

Answer 9

ooh.

Answer 10

cwrw is correct.

Answer 11

a bit tedious that...

Answer 12

it is isn't it? There has to be a simpler way...

Answer 13

Yeah, I want to believe there is a more though oriented way, instead of guess and check.

Answer 14

I got the answer btw <.< totally guess and check though.

Answer 15

The multiple is 122112, its \(64\cdot 1908\)

Answer 16

345658

Answer 17

345658*64= 22122112

Answer 18

I just realized, because there can only be 1's and 2's in the result, and its 6 digits long, there are only \(2^6\) possible candidates.

Answer 19

Take away the fact that 1 can't be the units digit, now there are \(2^5\) possible candidates.

Answer 20

Well, firstly, if a number is divisible by 64, it is divisible by 8.

If a number is divisible by 8, the last three digits should be divisible by 8.

Firstly, it must end with 2.

So the number is in the form ......2

so the last three digits is either 112, 122, 212, 222

Only 112 is divisible by 8.

So now we know the number is of the form ...112

Moreover, the number should also be divisible by 16. Now the last 4 digits have to be divisible by 16.

So it can either be ..2112 or ..1121

Only 2112 is divisible by 16.

So our number is ..2112

Now just test the last 4 possibilities (either 112112, 122112, 212112, 222112)

Answer 21

Very interesteing. :) You could keep going with 32 and 64, then you would be done.

Answer 22

Sorry, the fourth last line should read: So it can either be ..2112 or ..1112

Answer 23

great stuff

Answer 24

112112 doesnt work

Answer 25

122112 does work though.

Answer 26

its 122112. Because 22112 is divisible by 32, but 12112 isnt.

Answer 27

it does! yay!

Answer 28

Thanks everyone!

Answer 29

I wish i could give medals to alll of you XD

Answer 30

A reason why a 4 or higher digit number is divisible by 8 iff the last 3 digits is divisible by 8:

any four digit number (or higher) can be written as axyz

Now, axyz can be written as 1000a + xyz

We know 1000 is divisible by 8. So if axyz = 1000a + xyz is divisible by 8, it implies xyz is divisible by 8. This follows, because if any terms which are divisible by a number are added, the sum is also divisible by that number.