Six Consecutive Integers

Relevant wiki: Setting Up Equations

Let the consecutive integers be $n, n+1, n+2, n+3, n+4$ and $n+5$

The sum of these numbers is $6n+15$ .

The sum of the five integers will be $5n+15-p$ where $p$ is $0, 1, 2 ,3 , 4$ or $5$ .

$5n+15=2016+p$ , therefore $2016+p$ is divisible by $5$ and thus $p=4$ .

$5n+15=2020$

$n=401$ and $n+4=405$ .

The average of six consecutive numbers lies precisely between the third and fourth value.

Removing the first number will increase the average to be equal to the fourth value; removing the last number will decrease the average to be equal to the third value. Removing any of the other numbers will create an intermediate situation.

In this case, the average of the five remaining numbers is $2016 / 5 = 403\tfrac15$ .

Thus we know immediately that the third and fourth number are $403$ and $404$ , and the average of the original six numbers was $403\tfrac12$ .

The missing number is therefore $6\cdot 403\tfrac12 - 2016 = 2421 - 2016 = \boxed{405}.$

I dont know any fancy ways to solve this, but i still got it right by trial and error. First, I found the average of the 5 numbers by dividing 2016 by 5. (2016/5=403.2) Then I took the numbers 401, 402, 403, 404, 405, and 406 until i found 5 that sum equaled 2016. Those numbers were 401, 402, 403, 404, and 406. Therefore the number in the sequence that was missing was 405.

P.S. I made an educated guess to get that those weere the 6 consecutive numbers. I knew because of the average that 403 was either the 3rd or 4th number.

• 2000/5=400 leaves me with a simpler answer to find, so I ignored 2016 and used only the 16 at first
• As a first trial: 1+2+3+4+5+6=21
• 21-5=16 means 5 is our erased number
• But we need to add back in the 400.
• 405 was the erased number 😊

This is fundamentally the same as Paul Fournier's solution, but without much use of variables (which I should've done to save time but then I stumbled upon a bit more convoluted reasoning to arrive at the answer).

Any sum of 6 consecutive integers should leave a remainder of 3 when divided by 6. However, since 2016 is divisible by 6, the removed integer must be odd and divisible by 3 (if it's even it would be divisible by 6).

Any sum of 5 consecutive integers is divisible by 5. But since 2016 leaves a remainder of 1 when divided by 5, the sum of the first 5 written integers (before erasing) was 2015 and the sum of the last 5 (before erasing) was 2020.

Notice that for every 5 consecutive integers, the middle term is the average of the five. Therefore, the third smallest integer written by the professor $\frac{2015}{5}$ = 403. Thus, the 6 integers are 401, 402, 403, 404, 405, and 406. Only 405 is odd and divisible by 3, so 405 is the erased integer.

Let the numbers be N-2, N-1, N, N+1, N+2, N+3

Sum = 6N+3

Now if we remove one number, we will have a sum in the form "5N+x"

2016 = 5N+1 --------------------------- (1)

So x = 1

It means the number removed is = 6N+3 - (5N+1) = N+2

From (1), we have N = 403

Hence, the number removed = N+2 = 405

I attach a code in python :) sorry but this time I felt lazy

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for i in range(1,1200):
     sum = 0
     for j in range(i,i+6):
         sum += j
     for k in range(i,i+6):
             if (sum-k == 2016):
                 a = i
                 b = k
                 print "Starting number: ", i
                 print "Erased number: ", k
                 break