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' Tis Whole

Stage: 4 and 5 Challenge Level: Challenge Level:2 Challenge Level:2

Well done Daniel and Mariana from St Julian's School; Eduardo from The British School, Manila; Wesley from Milpitas High School; Chuyi Yang from Loughborough High School and Andrei from Tudor Vianu National College, Bucharest.

Trial and improvement was the chosen method for some of you, Mariana found all the solutions using inequalities and Andrei used a similar method carefully working through all the possibilities.

Removing 3 consecutive numbers from n numbers the solutions are: n=11 (removing 1,2 and 3); n=15 (removing 9, 10 and 11) and n=17 (removing 15, 16 and 17). If the numbers removed are not consecutive then the solutions are n=13 (where any three numbers adding up to 16 can be removed); n=15 (where any three numbers adding up to 30 can be removed and n= 17 (where any three numbers adding up to 48 can be removed).

Several of you noted that n is not even because if it was then (n-3) would be odd and 7.5 multiplied by (n-3) would not be a whole number, but as 7.5(n-3) is equal to the sum of whole numbers it has to be a whole number.

Chuyi, who is much younger than any of the other contributors, beautifully explained a trial and improvement method as follows showing that logical thinking can sometimes go a long way without needing a lot of mathematical knowledge.

Trial for n Trial Error
21 - This is just trying a high odd number
If n is 21 then x/18=7.5 so x will be 135. The sum of all the numbers between 1 and21 is 231 so the three removed numbers must total 96. There are such numberswhich are 31, 32 and 33 but since the numbersonly go up to 21 this is not possible.
This does not work because the numbers are out of range, so it means that this number is too high.
19 '?? This is going down by one odd number. If n is 19 then x/16=7.5 so x will be 120. The sum of all the numbers between 1 and 19 is 190 so the three removed numbers must total 70. But there are no three consecutive numbers that total 70 so 19 cannot be a possible value for n. This does not work, the three removed consecutive numbers will be around 23 and even that is out of range and too high.
17 '?? Once more going down by an odd number, is this low enough? If n is 17 then x/14=7.5 so x will be 105. The sum of all the numbers between 1 and 17 is 153 so the three removed numbers must total 48. There are such numbers which are 15, 16 and 17. This works well and therefore is a right answer. So this is just low enough as one of the removed numbers is the highest in the range.
15 '?? Even though a correct answer has been found already it always good to check if there is more than one answer. This was also the first estimate made so this is also to check the original estimate. If n is 15 then x/12=7.5 so x will be 90. The sum of all the numbers between 1 and 15 is 120 so the three removed numbers must total 30. There are such numbers which are 9, 10 and 11. This works well also and therefore is a right answer, so the first estimate made was correct
13 '?? This is going down now and I have realised that the total of the three consecutive numbers are becoming smaller and smaller and this is just experimenting on how low it can go If n is 13 then x/10=7.5 so x will be 75. The sum of all the numbers between 1 and 13 is 91 so the three removed numbers must total 16. There are no possible numbers so 13 is an impossible value for n. This is an impossible possibility for n. The rough number for the three consecutive numbers would be roughly five so it is still possible to go down further.
11 '?? This is going down further to see if there might be any possible values for n which are quite small. If n is 11 then x/8=7.5 so x will be 60. The sum of all the numbers between 1 and 11 is 66 so the three removed numbers must total 6. There are such numbers and they are 1, 2 and 3. This works well also and therefore is a right answer. Because this goes down to 1 as a removed number then it is impossible to go any further as 1 is the smallest value. Therefore 11 is the smallest possible answer for n.
Conclusion: n=17, 15 and 11
are the only possible solutions.

Here is Mariana's solution

1.Solution with consecutive numbers .

The sum of integers 1 to n is: $1 + 2 + 3 + \cdots + n = {n(n+1)\over 2}.$ Let the three numbers removed be $x$, $x+1$ and $x+2$ which add up to $3x+3$. Then $${{n(n+1)\over 2} -(3x+3)\over n-3}=7.5 \quad (1)$$ This simplifies to: $${n^2-14n+39\over 6} = x. \quad (2)$$ The smallest value of the first of the three consecutive numbers is 1 and the largest value is $(n-2)$. Therefore $1\leq x \leq n-2.$ Combining equation (2) with this inequality: $$1 \leq {n^2 - 14n +39\over 6} \leq n-2.$$ This simplifies to $6 \leq n^2-14n+39 \leq 6n-12.$ So $$0\leq n^2 -14n + 33 =(n - 11)(n - 3)$$ and $$n^2-20n + 51 =(n-3)(n-17)\leq 0.$$ From the first inequality $n\leq 3$ or $n\geq 11$ and from the second inequality $3\leq n \leq 17.$ However, $n=3$ is not a valid solution because, after removing three consecutive numbers, there would not be any numbers left to yield an average of 7.5. Therefore, $11\leq n \leq 17.$

Using equation (2) again, the following table contains the values of $x$ for all integer values of $n$ between 11 and 17.

n 11 12 13 14 15 16 17
x 1 2.5 4.3 6.5 9 11.83 15

Since $x$ has to be a whole number, then the only values of $n$ that are valid are $n=11$ (removing 1,2 and 3); $n=15$ (removing 9, 10 and 11) and $n=17$ (removing 15, 16 and 17).

2. Solution with non-consecutive numbers

Let the sum of the three numbers removed be $S$, then from equation (1): $${n(n+1)\over 2} - S = 7.5(n-3)$$ which simplifies to $${n^2 - 14n + 45 \over 2} = S.\quad (3)$$ The smallest value of $S$ is 6, when the numbers removed are 1, 2 and 3. The largest value of $S$ is $n + (n-1) + (n-2) = (3n - 3)$. Therefore $6\leq S \ leq 3n-3$. Combining this inequality with equation (3): $$6 \leq {n^2 - 14n + 45 \over 2} \leq 3n-3.$$ This reduces to $0\leq n^2 - 14n + 33 = (n-11)(n-3)$ and $n^2-20n+51 = (n-3)n-17)\leq 0.$

So, as in the first part $11\leq n \leq 17.$ The following table shows the values of S for all integer values of $n$ from 11 to 17.

n 11 12 13 14 15 16 17
S 6 10.5 16 22.5 30 38.5 48

As $S$ has to be a whole number, then $n$ can only take the values 11, 13, 15 and 17 giving the only possible solutions.

For $n=11$ the only three numbers adding up to 6 are 1, 2 and 3 which are consecutive and this solution came up in the first part.

For $n=13$ any three numbers adding up to 16 can be removed; for n=15 any three numbers adding up to 30 can be removed and for n= 17 any three numbers adding up to 48 can be removed.