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Answer: 52
Clare's method
Interchanging the digits should make the number smaller (about half), so the first digit should be greater (about double), like 42, 84 (only such pairs since Alberta's age should be even).
42, 24 +1 is a bit too big so try making 42 a bit bigger
52, 25 +1 = 26, perfect!
84, 48 +1 = 49 is half of 98, much too big
94, 49 +1 = 50 no
Charlie's method
If Alberta is $ab$ years old, then her age is $10a+b$
If the digits are interchanged this becomes $ba$, so $10b+a+1$ is half of $10a+b$ $$\begin{align}\therefore 10a+b &=2(10b+a+1)\\
10a+b&=20b+2a+2\\
8a&=19b+2\end{align}$$
$\therefore a=5, b=2$, Alberta is $52$
Alberta's Age
Age 14 to 16
ShortChallenge Level 
Answer: 52
Clare's method
Interchanging the digits should make the number smaller (about half), so the first digit should be greater (about double), like 42, 84 (only such pairs since Alberta's age should be even).
42, 24 +1 is a bit too big so try making 42 a bit bigger
52, 25 +1 = 26, perfect!
84, 48 +1 = 49 is half of 98, much too big
94, 49 +1 = 50 no
Charlie's method
If Alberta is $ab$ years old, then her age is $10a+b$
If the digits are interchanged this becomes $ba$, so $10b+a+1$ is half of $10a+b$ $$\begin{align}\therefore 10a+b &=2(10b+a+1)\\
10a+b&=20b+2a+2\\
8a&=19b+2\end{align}$$
| $b$ | $19b+2$ | $a$ |
|---|---|---|
| 1 | 21 | Not a whole number |
| 2 | 40 | 5 |
| 3 | 59 | Not a whole number |
| 4 | 78 | Not a whole number |
| 5 | 97 | Greater than 10 |
$\therefore a=5, b=2$, Alberta is $52$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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