Question 1209046
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Before we dive into the problem, let's consider the task of adding the first n positive integers
1, 2, 3, ..., n-1, n
where n is a positive integer. 


S = sum of those integers
S = 1+2+3+...+(n-1)+n


Make a copy of the sequence and reverse it so we have
<table border = "1" cellpadding = "5"><tr><td>Original</td><td>1</td><td>2</td><td>3</td><td>...</td><td>n-1</td><td>n</td></tr><tr><td>Reversed</td><td>n</td><td>n-1</td><td>n-2</td><td>...</td><td>2</td><td>1</td></tr></table>
Add straight down. 
The resulting sum for each column is n+1.
There are n of these sums.
Adding n copies of (n+1) leads to n*(n+1)
This result is equal to 2*S because the sum of any entire row is S.


2*S = n*(n+1)
S = n*(n+1)/2
S = 0.5*n*(n+1)


An example:
Let's add the consecutive integers from 1 to 10.
0.5*n*(n+1) = 0.5*10*(10+1) = 55
You can check that 1+2+3+4+5+6+7+8+9+10 = 55


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Now we can get to the problem your teacher asked.


Let x be the smallest integer that Pearl thinks of.
Then x+1 is next integer, and x+2 is after that, and so on until reaching x+6 as the seventh consecutive integer.


Summing those items gives
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)
= (x+x+x+x+x+x+x)+(1+2+3+4+5+6)
= 7x+(1+2+3+4+5+6)
= 7x+0.5*6*(6+1) ............... use the formula mentioned in the previous section
= 7x+21


We are then told that "<font color=blue>The sum of the integers is equal to 56 times the largest of the seven integers.</font>"
So,
sum = 56*(largest value)
7x+21 = 56*(x+6)


This is where we run into a problem. 
Solving this equation leads to x = -45/7 = -6.42857 approximately
The result is <u>not</u> an integer, so it seems like your teacher made a typo somewhere.
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