1989
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2011-02-01T00:00:01
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Can you draw a straight line across the centre of a clock face so that the numbers on both sides of the line have the same total?<br></br>
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<p style="text-align: center;"><mdo:image alt="clock face" height="185" src="ClockFace.gif" width="185"></mdo:image></p>
<p>Can you do this in another way?</p>
<p>Can you draw two lines (like the hands of the clock) to divide the clock face so that the total of the numbers on one side of the lines is twice the total on the other side? Can you do this in another way?</p>
<p>Can you divide the clock face so that that the total on one side of the lines is five times more than the total on the other side? Can you do this in another way?</p>
<p>Can you draw two lines to divide the numbers so that the total of the numbers on each side of the lines are both multiples of six? In how many different ways can you do this?</p>
<p>Can you draw two lines so that the numbers on each side add to a prime number? Can you do this in another way?</p>
<p>Can you find any other interesting ways to group the numbers on a clock face by drawing two lines?</p>
<p> </p>
<div>You might find <a href="/content/98/11/penta4/ClockFaces.pdf">this sheet</a> useful for working on.</div>
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<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6818&part=">Click here for a poster of this problem</a>.</div>
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<p class="editorial">1) Can you draw a straight line across the
centre of a clock face so that the numbers on both sides of the
line have the same total? Can you do this another way?</p>
<p class="editorial">Andy, Courtney, Lukas and others solved this
by looking at the sum of the numbers on the clock face, and then
trial and error. Lauren writes;</p>
By adding up all the numbers on the clock face, I got the total
$78$. I then divided $78$ by $2$ as we are dividing the clock face
into two equal parts. This equals $39$. Then by adding up six
consecutive numbers around the clock face I found that the line
should go between$9$ and $10$, and $3$ and $4$. As
$4+5+6+7+8+9=10+11+12+1+2+3=39$.<br></br>
<p class="editorial">A group from Archers Court in Dover solved the
problem using an algebraic method. This method also helps the
second question, on whether there is another solution.</p>
The sum of all the numbers on the clock face is $78$, so you want
to find six consecutive numbers that add up to $78\div2=39$. If we
call the first number $n$, then the sum of six consecutive numbers
is $n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=39$. By now gathering terms we
get the equation $6n+15=39$ So $6n=24$ and $n=4$ Then the six
consecutive numbers are $4,5,6,7,8,9$ and so the line runs from
after $9$ to after $3$. As there was only one solution to the
algebraic equation, then this is the only solution to the
problem.<br></br>
<p class="editorial">2) Can you draw two lines (like the hands of
the clock) to divide the clock face so that the total of the
numbers on one side of the lines is twice the total on the other
side? Can you do this in another way?</p>
<p class="editorial">Junnrui Hu wrote;</p>
I realized that the total should be divided intothree parts, so the
two of the parts together would then be twice as large as the other
part. $78\div3=26$. I then had to find consecutive numbers that
added to $26$. By choosing a number and adding up around the clock
from it until I reached $26$ or went over and missed $26$, I found
that $11+12+1+2=5+6+7+8=26$. So the line can go from between $10$
and $11$ to between $2$ and $3$ or from between $4$ and $5$ to
between $8$ and $9$.<br></br>
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<p class="editorial">3) Can you divide the clock face so that that
the total on one side of the lines is five times more than the
total on the other side? Can you do this in another way?</p>
<p class="editorial">Harry from the Beacon School solved this using
an algebraic method.</p>
The sum of all the numbers is $78$. If we call $x$ the total on the
small side, then $5x+x=78$, so $6x=78$ and $x=13$. So the sum on
the small side is $13$. Now we must find consecutive numbers that
add up to $13$. These are only $12+1$ and $6+7$. So there are two
places for the line, from after $11$ to after $1$ and from after
$5$ to after $7$.<br></br>
<p class="editorial">4) Can you draw two lines to divide the
numbers so that the total of the numbers on each side of the lines
are both multiples of six? In how many different ways can you do
this?</p>
<p class="editorial">Lily and Amy from Bow Brickhill School solved
this problem using a method of trial and error, by <span class="editorial">adding consecutive numbers until they totaled a
multiple of $6$. They found that $9+10+11+12=42 (7\times 6),
1+2+3=6 (1\times 6), 4+5+6+7+8=30 (5\times 6)$. Here they have
found three groups that sum to a multiple of six, so any
combination of these will w</span>ork, say having the lines after
$8$ and after $3$, or after $8$ and after $12$, or after $12$ and
after $3$.</p>
<p class="editorial">Susanna from St George's also used a trial and
error method to find several solutions.</p>
<p class="editorial">Junnrui Hu wrote;</p>
<div>To make both sums a multiple of six, I tried to find
consecutive sequences which added up to multiples of six. First I
used sequences of length $1$, which is just $12$ and $6$ on their
own. These can be isolated by the lines, and as $78$ is a multiple
of $6$, if you take away by isolating a multiple of $6$, the rest
will also be a multiple of $6$. Then looking at sequences of length
$2$ there are none as the sum is always odd. Trying three adjacent
terms there is a pattern $1+2+3=6=2\times 3=6\times 1,
2+3+4=9=3\times 3, 3+4+5=12=4\times 3=6\times 2$. So the groups
$(1,2,3), (3,4,5), (5,6,7), (7,8,9), (9,10,11)$ and $(11,12,1)$ all
are multiples of $6$.\</div>
<div>With 4 adjacent terms I found a pattern that starting on every
third number will work, so the groups $(3,4,5,6), (6,7,8,9),
(9,10,11,12)$ and $(12,1,2,3)$. In the same method I tried 5
consecutive numbers and found $(4,5,6,7,8)$ and $(10,11,12,1,2)$.
There are none with $6$ adjacent numbers as this sum is an odd
number as there are $3$ odd numbers in $6$ adjacent numbers.</div>
<div>And so you can find the solutions by isolating the groups,
$(6), (12), (1,2,3), (3,4,5),$ $(5,6,7), (7,8,9), (9,10,11),$
$(11,12,1), (3,4,5,6),$ $(6,7,8,9), (9,10,11,12),$ $(12,1,2,3),
(4,5,6,7,8)$ and $(10,11,12,1,2)$. So there are $14$ ways of making
to groups that add to multiples of $6$.</div>
<p class="editorial">5) Can you draw two lines so that the numbers
on each side add to a prime number? Can you do this in another
way?</p>
<p class="editorial">Junnrui Hu answers this;</p>
<div>I had to think of two odd prime numbers that added up to $78$
(since $2$ is the only even prime). First I wrote all the primes
from $1$ to $78$. $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73$. I found 7 pairs of primes that add
to $78$, $(5, 73), (11, 67), (17, 61), (19, 59), (31, 47), (37,
41)$. Now I look at ways of making the smaller number with
consecutive numbers. There are two ways of making $5 (5, 2+3)$, two
way for $7 (3+4, 7)$, two for $11 (11, 5+6)$, one for $17 (8+9)$,
one for $19 (9+10)$, and none for $31$ and $37$.</div>
<div>This gives $8$ different ways of diving up the clockface to
give two sides where the sums are prime numbers.</div>
<p class="editorial">Well done to Junnrui Hu for his excellent
answers, attempting every single question. Congratulations also to
Ian who solved every question using just his knowledge of
triangular and prime numbers.</p>
<p class="editorial">The last question remains open to anyone who
wishes to investigate it, can you find any other interesting ways
to group the numbers on a clock face by drawing two lines?</p>
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<h3><span style="font-weight: bold;">Why do this problem?</span></h3>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=1989&part=index">This problem</a> requires logical and creative thinking. It is also an intriguing and challenging way to practise addition and subtraction.</div>
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<h3>Possible approach</h3>
<div>A clock face without hands is a useful introduction for these challenges and it is a good idea to suggest that all the lines are drawn between the numbers rather than from number to number.</div>
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<div>Let learners have a go at the first part of the challenge in pairs without saying very much and then after a suitable length of time, bring them together to talk about how they went about finding a solution. Some might have used a trial and improvement approach; others may have found the total of all the numbers on the clock face first and then used some trial and improvement. Discussing
strategies at this stage will help children with the other parts of the problem.</div>
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<div>You can download <a href="/content/98/11/penta4/ClockFaces.pdf">this sheet</a> of clock faces to give to the class for rough working, jotting and recording.</div>
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<h3>Key questions</h3>
<div>What are the totals you are trying to get?</div>
<div>How do you know?</div>
What is the sum of all the numbers on the clock face?<br></br>
<h3>Possible extension</h3>
<div>Pupils could explore other divisions of the clock face and find the time on the clock when the hands do the dividing into different groups of numbers.</div>
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<h3>Possible support</h3>
<div>You could draw a line across a clock face and ask children to add the numbers on each side to show what is required. Having a calculator available would help if the addition is causing a problem in itself.</div>
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What are the totals you are trying to find in each case?<br></br>
How do you know?<br></br>
What is the sum of all the numbers on the clock
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From Jonathan at St Peter's College, Adelaide, Australia: <br></br>
"Dear Nrich, I have found a solution to Clock face. If you draw a
line from 12 to 6 and you go 12 minus 6 = 6. Every number greater
than 6 is like this. With any number 6 or under, you must add 6!
Cheerio!'' <br></br>
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Sakunthala has sent us a solution for the second half of the
problem: If you draw a line begining before 11 and ending before 3,
the total on one side of the line is half the other side.
Absolutely right. Well done!<br></br>
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Clock Face
Where can you draw a line on a clock face so that the numbers on
both sides have the same total?
Calculations and Numerical Methods
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Calculations and Numerical Methods
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