You may also like

problem icon

Pebbles

Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?

problem icon

Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

problem icon

GOT IT Now

For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?

14 Divisors

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

Mark from Beecroft Public School, Masturah from TES and Tarusha from Longthorpe Primary School sent in correct solutions for twelve divisors.

Laura from St Joseph's College, Ipswich gave correct answers for both twelve and fourteen divisors

$192$ is the lowest number with only $14$ factors.
I started with $12$ which had $6$ factors, then $24$ which had $8$ and I carried on until I got to $192$. The pairs are
$1$ and $192$
$2$ and $96$
$3$ and $64$
$4$ and $48$
$6$ and $32$
$8$ and $24$
$12$ and $16$
the lowest number with only twelve is $60$.
Pure trial and error lead me to this conclusion as it does not fit in with my twelve theory.

Raucher gave an excellent solution for a number with $12$ divisors.

$12$ can be written as $ 3\times 4$, $2\times6$, $ 2\times2\times3$ and $ 1 \times 12$ This gives a clue about how to use combinations of prime factors.

If the prime number divisors of a number $x$ are $a$ and $b$ then we can think about factors combinations of the two prime factors and $1$ based on the forms: ($1$, $a$, $a^2$)($1$, $b$, $b^2$, $b^3$) to give exactly $12$ divisors ($ 3\times 4$ combinations).
For example if $a=3$ and $b=2$ this gives the number $9 \times 8=72$ (one number with $12$ factors).

Similarly ($1$, $a$)($1$, $b$, $b^2$, $b^3$, $b^4$, $b^5$) to give exactly $12$ divisors ($ 2\times 6$ combinations).
If $a=3$ and $b=2$ this gives the number $3 \times 32=96$ (one number with $12$ factors).

Or ($1$, $a$)($1$, $b$)($1$, $c$, $c^2$) to give exactly $12$ divisors ($ 2\times2\times3$ combinations).
If $a=3$ and $b=5$ and $c=2$ this gives the number $3 \times 5 \times 4=60$ (the lowest number with $12$ factors).
The divisors are made from the $12$ combinations from $(1,3)(1,5)(1,2,4) $
$D60= {1, 5, 2, 4, 10, 20, 3, 15, 6, 12, 30, 60}$

Raucher's ideas can be developed into a general method for finding the answers, using prime factors:

Every whole number can be factorised into prime factors and here we are looking for a number like $(2^a)(3^b)(5^c)$ etc. Note that $2^a$ has $(1+a)$ factors so $2^{13}$ has 14 factors but could a number like $(2^a)(3^b)(5^c)$ be smaller?

Consider $(2^a)(3^b)$. This has $(1+a)(1+b)$ factors and so we want $(1+a)(1+b)=14$. Since $14=2 \times 7$ this must give $a=6$ and $b=1$ if we want to make the number as small as possible, and $2^{6}3^{1} = 192$ which is a lot smaller than $2^{13}$.

It remains to prove that it is impossible to find a smaller number which has more prime factors, say $(2^a)(3^b)(5^c)$, but this can't be done because 14 itself has only two factors. If we look for $a$, $b$ and $c$ such that $(1+a)(1+b)(1+c) = 14$ we find that $a$ or $b$ or $c$ must be zero.

Or, you could write a simple computer program!

A simple BASIC program might be something like this:

10 X=0
20 REPEAT
30 X=X+1
40 D=0
50 FOR Y = 1 TO X
60 R=X/Y-INT(X/Y)
70 IF R=0 THEN D=D+1
80 NEXT Y
90 UNTIL D=14
100 PRINT "Smallest number with exactly 14 divisors =";X

The above program is in BBC BASIC . If you are using QBASIC you will need to use a DO UNTIL LOOP in place of the REPEAT......UNTIL loop.