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# More divisibility

11 replies to this topic

### #1 bushindo

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Posted 12 February 2013 - 12:11 AM

This is an extension of Prime's excellent puzzle One's Divisibility

What is the smallest 101-plus-digit number consisting of only 4's, 5's, and 7's that is divisible by '7777.....777' (one hundred 7's)?

The solution *must contain* at least one 4, one 5, and one 7.

EDIT: clarified the properties of the required solution

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### #2 Rob_Gandy

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Posted 12 February 2013 - 12:29 AM

I'm assuming it has to have atleast one 4, 5 and 7?

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### #3 bushindo

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Posted 12 February 2013 - 02:52 AM

I'm assuming it has to have atleast one 4, 5 and 7?

That is correct. I'll edit the OP to clarify that point. Thanks.

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### #4 Prime

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Posted 15 February 2013 - 03:44 AM

This looks harder than the puzzle I posted.

Spoiler for Can't find it

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Past prime, actually.

### #5 Prime

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Posted 15 February 2013 - 04:01 AM

Found a roundabout way to display formatted text inside a BD post. Same thing as above, but visible.

Spoiler for can't find it, but here is a way

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Past prime, actually.

### #6 bushindo

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Posted 15 February 2013 - 04:59 AM

Found a roundabout way to display formatted text inside a BD post. Same thing as above, but visible.

Spoiler for can't find it, but here is a way

You are approaching this from the correct angle
Spoiler for

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### #7 Prime

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Posted 16 February 2013 - 02:43 AM

Spoiler for here's one big number

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Past prime, actually.

### #8 bushindo

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Posted 19 February 2013 - 07:17 AM

Spoiler for here s one big number

Excellent work so far. Sorry for not responding earlier. I was on a vacation and didn't have access to a computer
Spoiler for

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### #9 bushindo

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Posted 09 March 2013 - 06:54 AM

Spoiler for here s one big number

Excellent work so far. Sorry for not responding earlier. I was on a vacation and didn't have access to a computer
Spoiler for

Solution to the puzzle

Spoiler for

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### #10 Prime

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Posted 09 March 2013 - 08:16 AM

Indeed, this number is a bit smaller than the one I found.

Spoiler for another perspective

But where is the proof that it is the smallest number that meets  the conditions?

Edited by Prime, 09 March 2013 - 08:17 AM.

• 0

Past prime, actually.

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