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Cakes and Buns

Age 11 to 14 Short
Challenge Level

Answer: 9 cakes and 7 buns

Spotting patterns in the numbers
One cake + one bun = 23p + 39p = 62p

512 $\div$ 62 = 8 remainder 16, so buying 8 cakes and 8 buns costs 16p less

16 = 39 - 23 is the difference between them

So buy one more bun and one less cake
7 cakes and 9 buns


Using algebra
$23a$     $+$     $39b$     $=$     $512$
Those numbers are not very easy to work with so try writing it differently...

$23(a+b)$   $+\underbrace{16b}_{\text{multiple
of 16}}=\underbrace{512}_{\text{multiple
of 16}}$
So $(a+b)$ must be a multiple of $16$ as well.

Try $a+b=16$

$23\times16$      $+$     $16b$      $=$     $512$
     $ 368$       $ +$     $ 16b$      $ =$     $ 512$
$16b = 512 - 386 = 144$
$144=9\times16$ so $b=9$ and so $a=7$





This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.