480
/www/nrich/html/content/97/01/six1/
1
2011-02-01T00:00:01
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0">
<br/>
<p>The list below shows the first ten numbers together with their divisors (factors):</p>
<ol>
<li>$1$</li>
<li>$1$, $2$</li>
<li>$1$, $3$</li>
<li>$1$, $2$, $4$</li>
<li>$1$, $5$</li>
<li>$1$, $2$, $3$, $6$</li>
<li>$1$, $7$</li>
<li>$1$, $2$, $4$, $8$</li>
<li>$1$, $3$, $9$</li>
<li>$1$, $2$, $5$, $10$</li>
</ol>
<p>
What is the smallest number with exactly twelve divisors?
</p>
<p>
What is the smallest number with exactly fourteen divisors?
</p>
</mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<p><span class="editorial">Mark from Beecroft Public School,
Masturah from TES and Tarusha from Longthorpe Primary School sent
in correct solutions for twelve divisors.</span></p>
<p class="editorial">Laura from St Joseph's College, Ipswich gave
correct answers for both twelve and fourteen divisors</p>
<div>$192$ is the lowest number with only $14$ factors.</div>
<div>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</div>
<div>$1$ and $192$</div>
<div>$2$ and $96$</div>
<div>$3$ and $64$</div>
<div>$4$ and $48$</div>
<div>$6$ and $32$</div>
<div>$8$ and $24$</div>
<div>$12$ and $16$</div>
<div>the lowest number with only twelve is $60$.</div>
<div>Pure trial and error lead me to this conclusion as it does not
fit in with my twelve theory.</div>
<br></br>
<div class="editorial">Raucher gave an excellent solution for a
number with $12$ divisors.</div>
<br></br>
<div>$12$ 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.</div>
<br></br>
<div>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).</div>
<div>For example if $a=3$ and $b=2$ this gives the number $9 \times
8=72$ (one number with $12$ factors).</div>
<br></br>
<div>Similarly ($1$, $a$)($1$, $b$, $b^2$, $b^3$, $b^4$, $b^5$) to
give exactly $12$ divisors ($ 2\times 6$ combinations).</div>
<div>If $a=3$ and $b=2$ this gives the number $3 \times 32=96$ (one
number with $12$ factors).</div>
<br></br>
<div>Or ($1$, $a$)($1$, $b$)($1$, $c$, $c^2$) to give exactly $12$
divisors ($ 2\times2\times3$ combinations).</div>
<div>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).</div>
<div>The divisors are made from the $12$ combinations from
$(1,3)(1,5)(1,2,4) $</div>
<div>$D60= {1, 5, 2, 4, 10, 20, 3, 15, 6, 12, 30, 60}$</div>
<br></br>
<div class="editorial">Raucher's ideas can be developed into a
general method for finding the answers, using prime factors:</div>
<br></br>
<div>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?</div>
<p>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 $2^{13}$.</p>
<p>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.</p>
<p class="editorial"></p>
<p class="editorial">Or, you could write a simple computer
program!</p>
<p>A simple <strong>BASIC</strong> program might be something like
this:</p>
<code>10 X=0<br></br>
20 REPEAT<br></br>
30 X=X+1<br></br>
40 D=0<br></br>
50 FOR Y = 1 TO X<br></br>
60 R=X/Y-INT(X/Y)<br></br>
70 IF R=0 THEN D=D+1<br></br>
80 NEXT Y<br></br>
90 UNTIL D=14<br></br>
100 PRINT "Smallest number with exactly 14 divisors =";X</code>
<p>The above program is in <strong>BBC BASIC</strong> . If you are
using <strong>QBASIC</strong> you will need to use a <code>DO UNTIL
LOOP</code> in place of the <code>REPEAT......UNTIL</code>
loop.</p>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<h3><span style="font-weight: bold;">Why do</span> <span style="font-weight: bold;">this problem</span> <span style="font-weight: bold;">?</span></h3>
<p><a href="http://nrich.maths.org/public/viewer.php?obj_id=480">This problem</a> allows students to explore prime factorisation.</p>
<h3>Possible approach</h3>
<div>Ask for the factors of $12$. Ask for a number with more factors, and list them. Ask for a number with fewer factors. Ask for a bigger number with fewer factors, etc, etc, until the group has had enough revision of techniques to list factors systematically omitting none.</div>
<div>Challenge pupils to see who'll be first to find a number with exactly fourteen factors - but ensure they know that intermediate results should also be clear in their books, in case they're useful later.</div>
<div>Part-way through the lesson, get a progress report, ask pupils to describe useful approaches (without giving too much away).</div>
<div>Once a solution has been found, refine the challenge - to find the SMALLEST number with fourteen factors, and to provide a convincing argument that it is the smallest.</div>
<h3>Key questions</h3>
<div>Are there any more factors? How do you know?</div>
<div>"Big numbers have lots of factors" - discuss.</div>
<div>"That's odd, they all have an even numbers of factors - or do they?"</div>
<h3>Possible extension</h3>
<div>Find more numbers with fourteen factors, until you can describe/define ALL the numbers with fourteen factors.</div>
<div>Looking at the structure of prime factorisiation, and working with the primes and their powers as values to be varied may help pupils appreciate the role of prime factors and their powers.</div>
<br></br>
<div>Experiment until you can make an educated guess of - the modal number of factors for the numbers $1$-$10$, $1$-$100$, $501$-$600$, $a$-$b$.</div>
<br></br>
<div>Go and research the distribution of primes.</div>
<h3>Possible support</h3>
<div>Use ten rather than fourteen.</div>
<div>How many rectangles can be made with an area of $24$? - draw or cut out all the (integer) possibilities.</div>
<div>Shall we consider $4 \times 6$ the same as or different to $6 \times4$?</div>
<div>What other areas could we try? Everyone tries out one or two different areas - trying to get ALL the rectangles.</div>
<div>I want to find an area for which there are five different rectangles (or ten if we decide that $4 \times6$ is different to $6 \times4$).</div>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0">
<p> </p><p>Factorising numbers $1 , 2 , 3 , 4 , \dots$ until you find the first one with 14 divisors will give the answer here, which is $192$, but there is a better way! You could write a simple computer program ... but there is a more mathematical way! </p><p>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? </p><p>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 $2^{13}$. </p><p>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. </p> <p>
A simple
<strong>BASIC</strong>
program might be something like this:
</p>
<code>
10 X=0
<br/>
20 REPEAT
<br/>
30 X=X+1
<br/>
40 D=0
<br/>
50 FOR Y = 1 TO X
<br/>
60 R=X/Y-INT(X/Y)
<br/>
70 IF R=0 THEN D=D+1
<br/>
80 NEXT Y
<br/>
90 UNTIL D=14
<br/>
100 PRINT "Smallest number with exactly 14 divisors =";X
</code>
<p>
The above program is in
<strong>BBC BASIC</strong>
. If you are using
<strong>QBASIC</strong>
you will need to use a <code>DO UNTIL LOOP</code> in place of the <code>REPEAT......UNTIL</code> loop.
</p>
<p>Correct solutions to this problem were submitted by</p>
<p>
<strong>Rachel Took</strong>
,
<strong>Mary Egan</strong>
and
<strong>Rose Paul</strong>
, West Flegg Middle School
</p>
<p>
<strong>Ruth Bradman</strong>
, Lynn Grove High School
</p>
<p>
<strong>Hannah Stow</strong>
,
<strong>Lucy Hannah</strong>
,
<strong>Lawrence Tardelier</strong>
,
<br/>
<strong>Sarah Gray</strong>
and
<strong>Sarah Hannah</strong>
, Archbishop Sancroft High School
</p>
<p>
<strong>Wymondham High School</strong>
</p>
<br/>
</mdoxml>
2
4
0
0
1
0
0
14 divisors
What is the smallest number with exactly 14 divisors?
Numbers and the Number System
Properties of numbers
Numbers and the Number System
Divisibility
Numbers and the Number System
Factors and multiples
Information and Communications Technology
smartphone
Secondary Mapping Document
Factors, multiples and primes
Secondary Mapping Document
DisplayCabinet