Thank you for all the solutions you sent to this problem. Akash from South Island School used clear reasoning to reach an answer:
If the number's digits add up to an even number, then the digits will be either odd & odd or even & even.
If the ones digit is twice as much as the tens digit, then both the digits must be even. This is because of the fact that if an odd digit is multiplied by $2$, the product is always even.
Till now: Solution has even & even digits. The number is smaller than $25$. This means that the first digit has to be $2$. If the first digit is $2$, $2\times2=4$.
So the answer is $24$.
By the way, the number could be $00$ as well if it is counted.
Katarina, and Syed from Foxford School and Community College, both used the same method. Syed says:
To solve this problem I go through one clue at a time and use my elimination skill to find out what the number is:
Clue 1 : 'I am less than $25$'
So from this clue I list all the numbers below $25$.
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24$
Clue 2 : 'My unit digit is twice my tens digit'
This clue can be split into smaller clues:
'It has a tens and unit digit'
So now I get rid of all the numbers from the list from clue 1 that don't apply to this clue. So I have:
$10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24$
'Its unit digit is twice its tens digit'
Clue 3: 'My digits add up to an even number'
$1 + 2 = 3$ which not an even number so that it can only be:
Alexis approached the problem slightly differently. He wrote:
Many of you used Alexis' method, including Alex and Phoebe from Newton Poppleford Primary. Finally, Michaela from St Michael's C of E Primary had another way of solving the problem:
But why did you only write down the even numbers to start with, Michaela? How did you know the answer was even?
Well done all of you for explaining your reasoning so clearly.