### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### 14 Divisors

What is the smallest number with exactly 14 divisors?

### Summing Consecutive Numbers

Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?

# Weekly Problem 13 - 2011

##### Stage: 2 and 3 Short Challenge Level:

It is clear that each of a, b and c must be less than or equal to $10$. A brief inspection will show that the only combination of different square numbers which total $121$ is $81+36+4$.

More formally, the problem can be analysed by considering the remainders after dividing the square numbers less than $121$ $(1,4,9,16,25,36,49,64,81$ and $100)$ by three. The remainders are $(1,1,0,1,1,0,1,1,0$ and $1)$.

When $121$ is divided by $3$, the remainder is $1$. Therefore $a^2+b^2+c^2$ must also leave a remainder of $1$.

Now we can deduce that two of the three squares must leave a remainder of $0$ and so be multiples of $3$. There are three square numbers below $121$ which are multiples of three: $9, 36$ and $81$. Checking these, we see that $81$ and $36$ are the only pair to have a sum which differs from $121$ by a perfect square, namely $4$.

So $a+b+c=9+6+2=17$.

This problem is taken from the UKMT Mathematical Challenges.

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