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Counting Factors

Is there an efficient way to work out how many factors a large number has?

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Repeaters

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

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Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

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Age 11 to 14 Short Challenge Level:

Answer         or         


This solution is from Isobel and Rebekah from Wycombe High School in the UK. More solutions are shown below.
 


Well done to everyone who sent in a correct solution. Many of you appeared to use trial methods, like Allan, Maysa, Soyeon and Vaneeza from Greenacre Public School in Australia, who sent in their first trial:

We tried the problem out with trial and improvement and it turned out something like this:
         

3 wasn't a [multiple] of 2 but 6 was so we switched 6 with 3 and then it looked like this:
         

And it worked! 5 was a factor of 5, 1 was a factor of 5 and 3, 3 was a factor of 3 and 6, 2 was a factor of 6 and 4, and 4 was a factor of 4!

Ilya and Brodie from Grange Primary School in Australia said:
We found out that we could flip the numbers and it didn't make any difference.
So           is also a solution.


Inés from King's College of Alicante in Spain, and Moineraza from K12 International School in Dubai both started with the threes. Inés said:
For the number 3, I saw that the red 3 was a factor of the blue 3, also, I saw that the red 3 was a factor of the blue 6 and that the red 1 was a factor of the blue 3.
So, I plotted those numbers (   ) in the middle.
Then, the red 1 is a factor of the blue 5, and the red 5 is a factor of the blue 5 [which gives the sequence       ].
Finally, I realised that the red 2 was a factor of the blue 6, that the red 2 was a factor of the blue 4 and that the red 4 was a factor of the blue 4 [which gives          ].


Simon from Sefton Park School and Georgina from Parmiter's School in the UK and Billie and Lewis from Grange Primary School in Australia started with the one. This is Billie and Lewis' working:
We knew the 1 had to go next to a 5 and 3, because they are only divisible by themselves and 1.
And this is why the 5 went next to the 5 and the 3 went next to the 3 (    ).
We knew no other numbers were divisible by 5 so we put the 5s on the end next to the 1 and we knew 6 was divisible by 3 so we put the 6 next to the 3.
The only other number that went into 6 was 2 so we placed the 2 next to the 6 (      ).
That left us with two 4s so we placed them next to the 2 as we knew they were divisible by 2 and themselves.         

Meghna from The British School of Kathmandu, Maanya and Eliza from Stephen Perse Foundation Senior School in the UK and Spectrum Class at Blackman Elementary School in the USA all started at one end, beginning with either the fours or the fives. Spectrum Class described what they did:
First, we figured out that the red cards need to go on the outside edges since there are 5 red cards and only 4 blue cards.
Next, we made a list of the multiples of each number from 1-6.
It was evident that the red 5 would need to be on the end since the only factor that could be next to it was the number 1.
Since 1 is a factor of 5 and 3, the red 1 could go between those numbers.
We were able to put red 3 between blue 3 and blue 6 and red 2 between blue 6 and blue 4.
Red 4 went at the other end (        ).


Adithya from Hymers College, Pari from Leicester Grammar School, Nafisah and Julia from Stephen Perse Foundation School, Fred from The Latymer School, and Callum from Tudor Grange Academy Worcester, all in the UK, started by placing the fours and the fives at opposite ends. Julia said:
I realised that 5 and 5 and 4 and 4 were only factors of one number, each other. This meant that they either had to be at the start of the sequence or at the end of the sequence.
From there, Fred said:
The red card adjacent to the blue 5 must be a factor of 5, and only 1 of the remaining red cards is a factor of 5 - 1. The card adjacent to the blue 4 must therefore be a 2, as it is the only remaining factor of 4. Hence, the red card in the centre must be a 3, so we have:
         

All that we now need to do is insert the remaining 2 blue cards: 6 and 3.
The 3 must go between the 3 and the 1, leaving the 6 to go between the 3 and the 2:
         

You can find more short problems, arranged by curriculum topic, in our short problems collection.