Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Number theory

Find the lowest multiple of 84 which in base 10 comprises only of 6's and 7's. Repeat for 88.

Could someone please post solutions to both? Thanks.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,190

Re: Number theory

Tried some multiples but wasn't successful!
It is next to impossible to try all the multiples and look for a possible multiples that has only 6s and 7s in it.
Maybe, brute force!
Else the problem involves some cunning technique which eludes all of us solvers

---

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Number theory

Oh well, thanks for trying anyway

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Number theory

Well, for something to be a multiple of 84, it has to be a multiple of 4.
For that to happen, its last two digits must divide by 4, and so the last two digits of you number have to be 76. (Same goes for the 88 problem)

It's a start, anyway. You can apply more restrictions by realising that 84 divides by 3 and 88 divides by 11.

---

Why did the vector cross the road?
It wanted to be normal.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Number theory

Success!

Continuing on from my reasonning up there, you can use the fact that 84 is divisible by 3 to say that the magic number has to have an amount of 7's that is a multiple of 3. It also ends in 76, which means it needs at least 3 7's.

From there it's just a tiny bit of brute force.

7776 - NO
67776 - NO
76776 - YES.

914*84 = 76776.

---

The second one needs absolutely no brute force. It's a multiple of 8, meaning its last 3 digits are divisible by 8, meaning it ends in 776.
It's also divisible by 11, meaning that if you alternately add and subtract each of its digits, you'll end up with a multiple of 11.

We have ?776. Therefore, ?-7+7-6 = 11k.
Happily, this solves to give ? = 6, which fits the requirement.

So we have 6776. Checking shows that 77*88 = 6776. How pretty.

---

Why did the vector cross the road?
It wanted to be normal.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Number theory

Thanks mathsyperson!

It's a multiple of 8, meaning its last 3 digits are divisible by 8

I didn't know that, useful stuff

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Number theory

Yeah, there are tons of checks for various multiples, and they're (almost) all very useful.
The general version of that one is that a number is divisible by 2^n if its last n digits also are.

---

Why did the vector cross the road?
It wanted to be normal.