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A set of consecutive positive integers starting with 1 is written on the board. A student came along and erased one number. Average of remaining numbers is 61 15/20 . What was the number erased

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'15/20' isn't in simplest form; usually that would be written '3/4'. Normally I would think that was supposed to indicate that there were 20 numbers left being averaged, but that doesn't seem to apply here... –  Steven Stadnicki Sep 21 '12 at 1:46
    
@StevenStadnicki: I think it is a reasonable red herring for a problem like this. It makes one think the solution must involve a multiple of $20$ (at least it did me). Much better if there were an answer. –  Ross Millikan Sep 21 '12 at 4:05
    
Note that the answer that currently has most upvotes contains an error, as pointed out by Ross; if the error is corrected, the result is the same as in Ross' answer, namely that there is no solution. –  joriki Sep 21 '12 at 7:59
    
See this very similar question. The wording is very similar and a different average is used, and there is a solution. –  Peter Phipps Sep 21 '12 at 19:51
    
@PeterPhipps: But this one has an added twist. The method of the link doesn't work here, as N. S. and I prove. –  Ross Millikan Sep 22 '12 at 14:18
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3 Answers

up vote 4 down vote accepted

Let $n$ be the last number written. Lets say that $m$ is the erased number.

Then the sum of the numbers on the board is $\frac{n(n+1)}{2}-m$. Their average then is

$$\frac{\frac{n(n+1)}{2}-m}{n-1}=61 \frac{15}{20}$$

Multiplying by 2 you get

$$\frac{n(n+1)-2m}{n-1}=122\frac{3}{2}$$

$$\frac{n^2+n-2}{n-1}+\frac{2}{n-1}-\frac{2m}{n-1}=123\frac{1}{2}$$

$$n+2+\frac{2-2m}{n-1}=123 \frac{1}{2}.\tag{$*$}$$

Now, since $1 \leq m \leq n$ we have $$-2 \leq \frac{2-2m}{n-1} \leq 0 \,.$$

Using the fact that $n+2$ is an integer and $-2 \leq \frac{2-2m}{n-1} \leq 0 \,,$ in $(*)$, you see immediately that there are only two possibilities:

Case 1:

$n+2=124$ and $\frac{2-2m}{n-1}=-\frac{1}{2}$

Case 2:

$n+2=125$ and $\frac{2-2m}{n-1}=-\frac{3}{2}$

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Instead of considering case $1$ and case $2$, we could also make use of the fact that $n = 4k+1$ since $n(n+1)/2 - m = 61 \dfrac34 (n-1)$ –  user17762 Sep 21 '12 at 1:56
2  
But $61 \frac {15}{20} \times 2 = 122 \frac 32$ –  Ross Millikan Sep 21 '12 at 3:47
    
The logic of the solution goes through after the correction. –  Ross Millikan Sep 21 '12 at 12:45
    
@RossMillikan Thank you fixed. –  N. S. Sep 21 '12 at 13:59
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The average is $61 \frac 34$, and erasing one number can only move it by $\frac 12^+$, so the maximum had to be $122$ or $123$ But the denominator of the average is $4$, so the number of entries must be a multiple of $4$. This is a contradiction and there is no solution.

Added: There are $124$. The average starts out $62 \frac 12$, then the $3$ of $103$ was erased, changing it to $10.$ This reduces the sum by $93$ and the average by $\frac 34.$ Some will think this unfair, some will think it clever.

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If you ignore the the condition "starting with" the maxima program

for n:60 thru 130 do 
  for m:1 thru n-1 do 
    block([k], 
    k:(2*n^2-2*m^2-245*n+249*m)/4,
    if (integerp(k) and (m<=k) and (k<=n)) then
      print([m,k,n]));

will generate all solutions. So even if you interprete the condition "starting with" as "the first digit of the first number is 1" (I thought this is this kind of joke) you will have more than one solution.

[start,removed,end]

[60,63,64]
[58,64,66]
[56,65,68]
[54,66,70]
[52,67,72]
[50,68,74]
[48,69,76]
[46,70,78]
[44,71,80]
[42,72,82]
[40,73,84]
[38,74,86]
[36,75,88]
[34,76,90]
[32,77,92]
[30,78,94]
[28,79,96]
[26,80,98]
[24,81,100]
[22,82,102]
[20,83,104]
[18,84,106]
[16,85,108]
[14,86,110]
[12,87,112]
[10,88,114]
[8,89,116]
[6,90,118]
[4,91,120]
[2,92,122]
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