One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed?

Mathematics Asked by Drich1002 on January 3, 2022

One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $dfrac{163}4$. Which integer was removed?

Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1

I was hoping if someone could spot the flaw in my working for this question.


I began by letting the integer that was removed be $x$.

Then: $$frac{1 + 2 + cdots + (x-1) + (x+1) +cdots + n} {n-1} = frac{163}{4}$$

There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.

These are equal to $frac{x(x-1)}{2}$ and $frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:

$$frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = frac{163}{4}$$

which reduces to:

$$frac{n^2 + n – 2x}{2(n-1)} = frac {163}{4}$$

And then:

$$2(n^2 + n -2x) = 163(n-1)$$

At first I thought you could consider factors, as 163 was prime then:

$n-1 = 2$ giving $n = 3$ and $n^2 + n – 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn’t our positive integer.

I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help

4 Answers

Say you're removing $x$ from the set $1, 2, ldots, n$. The average of the resulting numbers will be at least $n/2$ (if you remove $n$) and at most $(n+2)/2$ (if you remove $1$.) So we have

$$ n/2 le 163/4 le (n+2)/2 $$

or, multiplying through by 4,

$$ 2n le 163 le 2n+2. $$.

So $n = 81$. You don't have to explicitly work out the number being removed, but it's $(1 + 2 ldots + 81) - (163/4) times 80$ = $(81 times 82)/2 - (163/4) times 80 = 61$.

Answered by Michael Lugo on January 3, 2022

The average of $1,2,3,ldots,n$ is the number halfway between the endpoints, $(n+1)/2,$ so the sum is $n(n+1)/2.$ Omitting $x$ from among $1,2,3,ldots,n,$ we get the sum $n(n+1)/2-x.$

Thus the average of $1,2,3,ldots,n$ must be a weighted average of $big( n(n+1)/2-xbig)/(n-1)$ and $x,$ with respective weights $(n-1)/n$ and $1/n.$ $$ frac{n-1} n left( frac{n(n+1)/2} {n-1} - frac x {n-1} right) + frac 1 ncdot x = frac{n+1} 2 $$ Therefore we have: begin{align} & frac{n-1} n cdot frac{163} 4 + frac x n = frac{n+1} 2 \[8pt] & frac{n(n+1)}{2(n-1)} - frac x {n-1} = frac{163} 4 end{align} So we get a system of two equations that is quadratic in $n$ and linear in $x.$ I'd try solving for one of those two in terms of the other and then substituting and solving the remaining equation.

Answered by Michael Hardy on January 3, 2022

This is not how you are supposed to solve it, but I feel like cheating. We have the equation $2(n^2+n-2x)=163(n-1)$, and $1leq xleq n$.

If you assume that $x=1$ then you solve for $n$ using the quadratic formula, and you obtain $79.5$.

If you assume that $x=n$ then you solve for $n$ you obtain $81.5$. Thus $n=80$ or $n=81$.

If $n=80$ then you can solve for $x$ and obtain $83=4x$, wrong. So $n=81$. Solving again yields $4n=244$, and ding ding, we have a winner.

As everyone else seems to have done this in the way I considered cheating, I should expand on the way I initially did it, which is completely different.

We have $2(n^2+n-2x)=163(n-1)$, and $1leq xleq n$. We see that $n$ is odd, so $n=2m+1$. Substituting in and cancelling the 2s yields

$$(2m+1)^2+(2m+1)-2x=163m$$ or $$4m^2+2-2x=157m.$$ Write $y=x-1$ and also, the LHS is even, so $m=2a$ is even. More substituting and removing the $2$ from both sides yields $$8a^2-y=157a.$$ The crucial point: we see that $amid y$. Since $n=2m+1=4a+1$, and $y$ is a multiple of $a$, $y=alpha a$ for $alpha$ between $1$ and $4$. Dividing through by $a$ yields $$8a-alpha=157.$$ Taking congruences modulo $8$ yields $alphaequiv 3bmod 8$, so $alpha=3$. Thus $y=3a$, so $x=3a+1$ and $n=4a+1$. We put this back into the top equation, $2(n^2+n-2x)=163(n-1)$, to obtain $a=20$, so $n=81$, $x=61$.

Answered by David A. Craven on January 3, 2022

We know that $n$ is odd. Notice that

$$ n^2 - n leq n^2 + n -2x leq n^2 +n - 2 $$

$$ implies n^2 - n leq frac{163}{2}(n-1) leq n^2 +n - 2 $$

which gives us $n geq 79.5$ and $nleq 81.5$, so $n=81$

Answered by Attila1177298 on January 3, 2022

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