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2011-02-01T00:00:01
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<span style="font-style: italic;">Please note: Although it involves
only elementary mathematical ideas, this problem is very difficult
and will probably take a long time to
solve!</span> <br></br>
<br></br>
<div style="float: left;"> </div>
<br></br>
A mathematician goes into a supermarket and buys four items.
<p>It has been a while since she has used a calculator and she
multiplies the cost (in pounds, using the decimal point for the
pence) instead of adding them.</p>
<p>At the checkout she says, "So that's £7.11" and the
checkout man, correctly adding the items, agrees.</p>
<p>Find four possible prices of the items.</p>
<br></br>
<span style="font-weight: bold;">Extension:</span> Prove that the
costs giving rise to £7.11 are unique.<br></br>
<br></br>
This problem is adapted from "Sums for Smart Kids" by Laurie
Buxton, published by <a href="http://www.beam.co.uk/">BEAM
Education</a> <br></br>
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<p class="editorial">This is a tricky problem to solve...</p>
<p class="editorial">Congratulations to Jeremy, formerly of Epsom College, who sent in his solution which involves some concise mathematical reasoning combined with a little trial and error:</p>
<div>Preamble: The "sum of the digits test" says that, if the sum of the digits of a whole number is divisble by 3, so is the number.</div>
<div>The sum of the 4 amounts (in pence) is 711. The sum of these digits is divisble by 3, so 3 divides 711 (leaving 237).</div>
<div>By the same reasoning, 237 is also divisible by 3 (leaving 79, a prime number).</div>
<br></br>
<div style="font-weight: bold;">So a trial solution is (in pounds) 1 x 3 x 3 x 0.79 (we need 4 factors).</div>
<br></br>
<div>However, this does not add up correctly (it adds to 779 pence - but it is close).</div>
<br></br>
<div style="font-weight: bold;">Now we need to multiply each of the amounts by a multiplier r1, r2,r3,r4 so that r1.r2.r3.r4 = 1.</div>
<br></br>
<div>The multipliers are simple fractions like 1/2 so that the new number is still a whole number of pence in each case.</div>
<div>For the first 3 digits (1,3,3) we will only get totals of 150 p, 100p, 50p, 25p, 20 p, 10 p etc when we apply the multiplier - yet the sum must end in a 1p.</div>
<br></br>
<div>So we will try r4 = 4 (ie. take 4 lots of.79), as this ends in a 6</div>
<div>(4 is the only multiple of 9 that does end in a 6 among the digits 1 - 9)</div>
<div>and when combined with a 25 p adjustment this gives a total ending in 1p.</div>
<br></br>
<div>The only multiple of 9 that ends in a 1 is 81 and 9 lots of 0.79 is too many to work, so this rules out the other adjustments to £1, £3 and £3 which are multiples of 10.</div>
<div> </div>
<div>(£1 x 5/4) x (£3 x 1/2) x (£3 x 2/5) x (£0.79 x 4) gives the correct answer:</div>
<div> </div>
<div><span style="font-weight: bold;">£1.25 x £1.50 x £1.20 x £3.16 = £7.11</span></div>
<div style="font-weight: bold;">£1.25 + £1.50 + £1.20 + £3.16 = £7.11</div>
<br></br>
<div class="editorial">Note that the product of the multipliers equals one: 5/4 x 1/2 x 2/5 x 4 = 1</div></mdoxml>
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<mdoxml version="1.0"><br></br>A challenging problem that requires appreciation of factors and multiples and a systematic approach - not for the faint-hearted!<br></br>
<br></br>
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<p>What are the factors of $711$?</p>
<p>If you have four numbers whose product is $7.11$, you can double
one of the numbers and halve one of the others, and the product
will still be $7.11$,</p>
<p>or</p>
<p>you can double one of the numbers, treble another and divide a
third by six, and the product will still be $7.11$,</p>
<p>or...</p>
<p>What else can you do that leaves the product unchanged?</p>
<p>This is quite a challenging problem. If you get part of the way
and would like more hints, try posting a question on the Please
Explain section of the <a href="http://nrich.maths.org/discus/messages/board-topics.html">AskNRICH</a>
web board.</p>
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<mdoxml version="1.0"><br></br>This is a very tricky problem and it has puzzled many of you for
several months. <br></br><br></br>However, Douglas at Burgoyne Middle School has
cracked it. <br></br><br></br>He says that he found a similar problem in a book which
actually gave three of the prices. This made it a bit easier to get
started.
<br></br><br></br>Douglas lists the answers as:
£1.20, £1.25, £1.50 and £3.16.
<br></br><br></br>Unfortunately, when we asked how he did it, Douglas couldn't
remember his method.
If you think that you can explain how to arrive at these prices,
please let us know.<br></br>
<br></br>
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Shopping Basket
A mathematician goes into a supermarket and buys four items. Using
a calculator she multiplies the cost instead of adding them. How
can her answer be the same as the total at the till?
Using, Applying and Reasoning about Mathematics
Working systematically
Numbers and the Number System
Factors and multiples
Fractions, Decimals, Percentages, Ratio and Proportion
Calculating with decimals
Fractions, Decimals, Percentages, Ratio and Proportion
Calculating with fractions