Puzzle

[Dave Moore]

There is an integer which is a multiple of 72.
Each of its digits is either 0 or 1.
What is the smallest possible number it could be?





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[Torn Rumero DeBrak]

Hint: 10....080 is a multiple of 72. ;-)

[Julius Kavay]

On 10.12.2020 10:45, Dave Moore wrote:
> There is an integer which is a multiple of 72.
> Each of its digits is either 0 or 1.
> What is the smallest possible number it could be?

1) the trivial solution: n = 0
2) you talk about INTEGER but say nothing about the number base (maybe
it's 10, maybe it's 2 or something else) I simply use base 16 (i.e. hex)
numbers:

23F048(hex) * 72(hex) = 10010010(hex)

Would you accept this solution?


Regards and have a nice day
Julius
--

[Basil Jet]

On 10/12/2020 09:45, Dave Moore wrote:
> There is an integer which is a multiple of 72.
> Each of its digits is either 0 or 1.
> What is the smallest possible number it could be?

1 = 1 mod 72
10 = 10 mod 72
100 = 28 mod 72
1000 = -8 mod 72
10000 = -8 mod 72
100000 = -8 mod 72

All larger powers of 10 give you -8 mod 72

A one in the least three significant digits will add a 1, 2, or 8 to the
total mod 8, which can never be cancelled out to get back to 0 mod 8, or
0 mod 72.

Therefore any number composed of zeroes and ones which is 0 mod 72 must
contain a number of ones which is a multiple of 9, all of which are in
the 1000's digit or greater.

The smallest compliant number (apart from zero) is 111,111,111,000.

--
Basil Jet recently enjoyed listening to
Steve Reich - Early Works

[Dave Moore]

"Basil Jet" <ba...@spamspamspam.com> wrote in message
news:rqvgar$3o5$1...@dont-email.me...

Correct solution.
Well done!

[Basil Jet]

On 11/12/2020 10:06, Basil Jet wrote:
> On 10/12/2020 09:45, Dave Moore wrote:
>> There is an integer which is a multiple of 72.
>> Each of its digits is either 0 or 1.
>> What is the smallest possible number it could be?
>
> 1 = 1 mod 72
> 10 = 10 mod 72
> 100 = 28 mod 72
> 1000 = -8 mod 72
> 10000 = -8 mod 72
> 100000 = -8 mod 72
>
> All larger powers of 10 give you -8 mod 72
>
> A one in the least three significant digits will add a 1, 2, or 8

Er, I meant a 1, 2 or 4

> to the
> total mod 8, which can never be cancelled out to get back to 0 mod 8, or
> 0 mod 72.
>
> Therefore any number composed of zeroes and ones which is 0 mod 72 must
> contain a number of ones which is a multiple of 9, all of which are in
> the 1000's digit or greater.
>
> The smallest compliant number (apart from zero) is 111,111,111,000.

--
Basil Jet recently enjoyed listening to

Steve Reich - 1998 - Music For 18 Musicians (1976)

[WiebeV]

Op 10-12-2020 om 10:45 schreef Dave Moore:

That integer can't end in a 1, so it has to be a multiple of 360 ( =5 * 72).

If we cut off the last digit, the question becomes "which multiple of 36
(360 without last zero) has only 1's and 0's?

Again, last digit can't be a 1, so we must look at multiples of 36 which
end in a zero. That means we have multiples of 180 (= 5*36).

And now again we drop a digit and the question becomes: "which multiple
of 18 has only 1's and 0's?

And again, last digit can't be a 1, so we must look at multiples of 18
which end in a zero. That means we have multiples of 90 (= 5*18).

And now again we drop a digit and the question becomes: "which multiple
of 9 has only 1's and 0's?

That answer is of course 111.111.111 (the smallest answer), because it
must be divisible by 9.

Since we dropped 3 zeros, the final answer is indeed (looking at Basil's
solution) 111.111.111.000

Solution of Basil was very nice, I've learnt something!

Wiebe Veldhuis

[Dave Moore]

"Dave Moore" <da...@djmoorenospam.fsnet.co.uk> wrote in message
news:rqsqmg$1tvk$1...@gioia.aioe.org...

> There is an integer which is a multiple of 72.
> Each of its digits is either 0 or 1.
> What is the smallest possible number it could be?
>
>
>

Several correct solutions so far.

Here is another way of solving the puzzle --

Since the required integer is a multiple of 72, it must be divisible by both
8 and 9.

To be divisible by 9, the digits must sum to 9 -- hence, for the smallest
case it has 9 ones.

To be divisible by 8, the last 3 digits must be divisible by 8.

The only possible last 3 digits are -

000
001
010
011
100
101
110
111

and the only one of those divisible by 8 is 000

Hence the required number is 111,111,111,000

[WiebeV]

Op 13-12-2020 om 22:12 schreef Dave Moore:

'Simple' solutions are the best!