Most maths teachers tend to be a bit sceptical when a pupil correctly solves a difficult problem very quickly. Honestly, our first instinct may be to wonder if they’ve hoodwinked us a little—maybe by using a calculator or looking up the answer online. And yes, that can be frustrating!
But every now and then, a child surprises us with a brilliant and completely valid shortcut—one that shows they truly understand the maths. Those moments… they’re worth celebrating!
This reminds me of one of my favorite maths stories about whether so-called ‘maths geniuses’ ever cheated. The story comes from the fascinating website mathematicalmysteries.com, and it’s too good not to share. One of the greatest mathematical minds of all time was once considered a possible cheat. It is not hard to see why.
When Carl Friedrich Gauss was just three years old (in 1780) he corrected a maths problem his father had previously got wrong. However, it was when he was just seven years old though that his genius led to a suspicion that he was a cheat. The story goes that his teacher decided to have an easy hour-long class, so asked the kids to use their blackboards to determine what the answer would be if you added 1+2+3+4… all the way for all numbers up to +100. At the end of the lesson most kids had scrawlings all over their blackboards, as the teacher anticipated. You can only imagine that most had the answer incorrect; it’s not easy to manually add up so many numbers as a seven year old without making an error.
But when the teacher got to Gauss’ blackboard there was only one number 5,050.
This was the correct answer. And a stunned teacher must have assumed some cheating, but the seven-year-old Gauss soon explained how he reached the answer: Instead of adding up the numbers in order he added them up from the ends. It doesn’t matter after all which order the numbers are added, it should give the same answer. So he added up 100+1, then 99+2, then 98+3 and as you will notice, if you add up all the pairs, you will have exactly 50 groups of 101, and 50 multiplied by 101 gives us 5,050. Genius!
Parent problem of the week
This week’s problem is a combination of English and maths:
What is the smallest whole number to contain all the vowels ‘a, e, i, o, u’ as well as the letter ‘y’ when written out in full? One proviso – the word ‘and’ does NOT count as part of the spelling of a number. So ‘one hundred and one’ does NOT count as having the letter ‘a’.
Solution to last week’s problem
Looking at the 3 sums below, what is the answer to the 4th one?
· 8 + 2 = 16106
· 5 + 4 = 2091
· 9 + 6 = 54153
· 7 + 3 = ?
The answer is 21104.
To arrive at each answer you first multiply the two numbers, then add the two numbers, and finally subtract the two numbers. Your final answer is each of these answers strung together.
Wishing you a great weekend.
Mr Pretorius, Head of Mathematics