### Counting Factors

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

### 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.

### Oh! Hidden Inside?

Find the number which has 8 divisors, such that the product of the divisors is 331776.

# Just Repeat

##### Stage: 3 Challenge Level:

A number of pleasing solutions were received.

Here is one offered by Andrei of School 205 Bucharest:

I observed that number 234234 is:

234234 = 234x1000 + 234 = 234(1000 + 1) = 1001 x 234

But 1001 is always divisible by 91:

1001 = 7 x 11 x 13 = 77 x 13 = 91 x 11 = 143 x 7

The same is true for 973973:

973973 = 1001 x 973

In fact, if you repeat any 3-digit number in that manner, you obtain a number divisible by 1001, so by 7, 11, 13, 77, 91 and 143:

Now, I look through other patters:

- if you repeat a 2-digit number twice,you obtain a number divisible by 101:

E.g. 65 ...6565

6565/101 = 65

- if you repeat a 2-digit numberthree times you obtain a number divisible by 3, 7, 13, 37 and all combinations obtained by multiplying them (21, 39, 111, 91, 259, 351, 273, and 3367).

- if you repeat a 4-digit number twice

Here, I obtained a prime number.

I did not continue further, as I would obtain too large numbers.