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Counting Factors

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Jaeyoon, Adithya, Himesh, Maaz, Andrew, Arun and Tom found a method to work out how many factors a number had. Maaz from Stepney Green Maths, Computing and Science College said:
You must add $1$ to each of the powers and then you must multiply [the results.]
For example $25725=5^2\times3^1\times7^3$
Factors of $25725:$  
$1\times 25725,$  
$3 \times 8575,$  
$5 \times 5145,$  
$7 \times 3675,$  
$15 \times 1715,$  
$21 \times 1225,$  
$25 \times 1029,$  
$35 \times 735,$  
$49 \times 525,$  
$75 \times 343,$  
$105 \times 245,$  
$147 \times 175$

There are $24$ in total, and $(2+1) \times (1+1) \times (3+1) = 24$


Tom from Litherland High School in the UK explained that
This [works] because the number of arrangements possible with its prime factors determines the number of factors. Add $1$ because $1$ is a possible multiple ($2^0 = 1$) (so for example if you have $2^2$, then $2^2$, $2^1$ and $2^0$ are possible factors)


Jaeyoon, Adithya, Himesh and Andrew used this approach to find the smallest numbers with 14 and 15 factors. For 14 factors, Andrew said:
First I found out all the factors that when multiplied would equal $14$.
$1 \times14$ and $2\times 7$
I then found the numbers which, raised to the original exponents, when
multiplied would give you the smallest answer.
$N=2^{13}\times3^0=8192$                                            $(13+1)(0+1)=14\times1=14$
$N = 2^1 \times 3^6=1458 $                                                $(1 + 1) ( 6+1) = 2 \times7=14$
or $N = 2^6 \times 3^1  = 192$

The smallest such number is $192$.


For 15 factors, Jaeyoon from South Korea said:
Let's first find the distinct sets of numbers that multiply to $15$, where every number is greater than $1:$ $(3, 5)$ and $(15).$ This means the numbers are in the form $a^2\times b^4,$
or $a^{14},$ where $a$ and $b$ are unique primes.

The smallest for $a^2\times b^4$ is $3^2\times2^4=144.$
$2^{14}=16384.$
Obviously, $144<16384.$
The smallest number with $15$ factors is $144.$


Jaeyoon, Adithya and Himesh all correctly found the smallest number with 18 factors. Adithya, from Hymers College in the UK, said:
$18= 18\times1$ or $9\times2$ or $6\times3$ or $2\times3\times3$
The first three prime numbers are $2,3$ and $5.$ Increasing the exponents of $2$ and $3$ gives me $2^2\times3^2\times5=180.$


Jaeyoon, Adithya, Himesh and Andrew found that square numbers have an odd number of factors. Adithya used Alison's method to explain why:
The numbers which have an odd number of factors are always square numbers
because you have to include the repeated factor of that number which is its
square root. For example, Factors of 36 are
1 and 36
2 and 18
3 and 12
4 and 9
6 and 6
The repeated factor of 6 will be omitted if we list the factors hence why
36 has 9 factors which is odd.


Himesh from Shenley Brook End School in the UK used Charlie's method to explain why:
Only square numbers have an odd number of factors.The prime factorization of a square number has to have only even numbers in the exponents and if you add 1 to an even it becomes an odd, odd$\times$odd$\times$odd$\times$odd.... = odd.


Himesh and Adithya both correctly completed both parts of the extension. This is Himesh's work:
Smallest number with $100$ factors:

$100=2\times2\times5\times5$
so $2^4\times3^4\times5\times7$ is the smallest number with $100$ divisors.

The number with the most factors which is below $1000$, by trial and
improvement you get $840$ which has $32$ factors. If we go over $32$ then the
number we chose has to be bigger than $1000.$


Nobody explained in detail how they had got the number less than 1000 with the most factors, but it can be done by considering small primes:
$2\times3\times 5 \times7=210<1000$ and $2\times3\times5\times7\times11=2310>1000$, so only primes less than $11$ should be used.
$210\times5=1050>1000$ and $210\times4=840<1000.$
So $2\times3\times5\times7$ can be multiplied by $2^2$ or $3^1$ to give a number less than $1000.$ It is sensible to multiply by $2^2$ as it has more factors, giving $2^3\times3\times5\times7=840,$ which has $4\times2\times2\times2=32$ factors.
Considering numbers which are products of only $2$s, $3$s and $5$s (or only $2$s and $3$s), this would change the number of factors, $4\times2\times2\times2$, by removing one of the $2$s and replacing the $4$ with a larger number. Removing one of the $2$s would leave $4\times2\times2=16$. $5\times2\times2=20,$ $6\times2\times2=24,$ $7\times2\times2=28,$ $8\times2\times2=32$ - so the power of $2$ would have to be at least $9$ to get a number with more than $32$ factors in this form.
However $2^9\times3\times5=7680>1000$
So $840$ is the number less than $1000$ with the most factors.