For this challenge, you'll need to play Got It! Can you explain the
strategy for winning this game with any target?
Draw a square. A second square of the same size slides around the
first always maintaining contact and keeping the same orientation.
How far does the dot travel?
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
Hannah from Millom School in Cumbria sent in a
nicely articulated solution:
The diagram shows that the sum of the first $4$ odd numbers is
$16$ because there are $4$ rows and $4$ columns of counters. For
the sum of the first $20$ odd numbers there are $20$ rows and $20$
columns. So if I do $30 \times 30$ (or $30$ squared) I get an
answer of $900$. For the sum of the first $60$ odd numbers there
are $60$ rows and $60$ columns. So if I do $60 \times 60$ (or $60$
squared) I get an answer of $3600$.
If you want the sum of the first $n$ odd numbers the answer
would be $n$ squared.
I worked out that $153$ is the $77$th odd number. I did this by
adding one (to get $154$) and then divided the answer by $2$.
The sum of the first $77$ odd numbers is $77\times77$ which is
To find $51 + 53+ 55+\ldots+ 149 + 151 + 153$ I used the answer
from the previous question which was $5929$.
As we were starting at $51$ this time and not $1$, I needed to
find the sum of all the odd numbers from $1$ up to $49$. I found
that $49$ is the $25$th odd number (by adding $1$ to $49$ and then
dividing the answer by $2$) So the sum of the odd numbers from $1$
to $49$ is $25$ squared which is $625$.
Finally I took $625$ away from $5929$ to give an answer of
David decided to use algebra to explain his
The sum of the first $30$ odd numbers $= 30^2 = 900$.
The sum of the first $60$ odd numbers $= 60^2 = 3600$
Quick Method: The sum of the first $n$ odd numbers $= n^2$
What is the sum of $1 + 3 + \ldots + 149 + 151 + 153$?
The formula for odd numbers is $2n-1$
We have: $2n-1 = 153$
$2n = 154$
$n = 77$
So $153$ is the $77$th odd number. The sum of the first $77$ odd
numbers $= 77^2 = 5929$. Therefore, the sum of $1 + 3 + \ldots+ 149
+ 151 + 153 = 5929$
What is the value of $51 + 53 + 55 + \ldots+ 149 + 151 +
The answer is the sum of ($1 + 3 + \ldots + 149 + 151 + 153$ -
which is already worked out) minus the sum of ($1 + 3 + \ldots
$49$ is the $25$th odd number (as $2n-1 = 49 \Rightarrow 2n =
50$, so $n = 25$)
Therefore the value of $51 + 53 + 55 + ... + 149 + 151 + 153 =
77^2 - 25^2 = 5304$
Ian from Colton Primary School and
Hannah from Thorner C of E Primary School also completed
particularly nice solutions, but we don't have space to show them