Challenge Level

Let $39=a+b$, $48=a+c$ and $51=b+c$.

$c$ is the largest number because $a+b=39$ is the smallest sum.

Adding together the sums containing $c$, $$\begin{align}48+51&=a+c+b+c\\

\Rightarrow 99&=\underbrace{a+b}_{39}+2c\\

99&=39+2c\\

\Rightarrow60&=2c\\

\Rightarrow30&=c\end{align}$$

Let $39=a+b$, $48=a+c$ and $51=b+c$.

$c$ is the largest number because $a+b=39$ is the smallest sum.

Subtracting $39=a+b$ from $48=a+c$ we can eliminate $a$:

$$\begin{align}48-39&=a+c-(a+b)\\

\Rightarrow 9&=c-b\end{align}$$

Adding $9=c-b$ and $51=b+c$ we can eliminate $b$:

$$\begin{align}9+51&=c-b+c+b\\

\Rightarrow60&=2c\\

\Rightarrow30&=c\end{align}$$

51 = largest number added to middle number.

51 = 25 + 26, so the largest number must be at least 26.

largest number | middle number (=51-largest number) | smallest number (=39- middle number) | smallest number + largest number (should be 48) |
---|---|---|---|

26 | 25 | 14 | 40 - much too small |

32 | 19 | 20 - too large | 52 - too large |

29 | 22 | 17 | 46 - too small |

30 | 21 | 18 | 48 |

39 is the smallest number so it must be the sum of the smallest and middle numbers.

51 is the largest number so it must be the the sum of the largest and middle numbers.

48 must be the sum of the largest and smallest numbers.

So when the smallest number is added to the middle and largest numbers, the sums are 39 and 48 respectively. So the difference between 39 and 48 must be the same as the difference between the middle and largest numbers. So the largest number must be 9 more than the middle number.

So 51 is the sum of two numbers whose difference is 9.

25 + 26 = 51 difference: 1

23 + 28 = 51 difference: 5

21 + 30 = 51 difference: 9

So the largest number is 30.

39 is the smallest number so it must be the sum of the smallest and middle numbers.

51 is the largest number so it must be the the sum of the largest and middle numbers.

48 must be the sum of the largest and smallest numbers.

We are interested in the largest number. Adding together the sums containing the largest number,

the largest number + the middle number + the largest number + the smallest number = 51 + 48 = 99.

But that is the largest number twice, plus the sum of the smallest and middle numbers. But we know that the sum of the smallest and middle numbers is 39! So the largest number twice is 99 $-$ 39 = 60.

So the largest number is 30.

You can find more short problems, arranged by curriculum topic, in our short problems collection.