Question 111238: the sum of seven consecutive integers is 980. How many of them are prime?
Found 3 solutions by josmiceli, dayne514, jgr45:
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! I cheated because I have a list of primes from 1 - 1,000.
I will consider 2 cases. First when the 1st number is odd
Then I look at when the 1st number is even.
To guarantee a number will be odd, call it
No matter what  is is odd.
The list is
(1) odd
(2) even 
(3) odd 
(3) even 
(4) odd 
(5) even 
(6) odd  add them up
 so, the list is:
137, 138, 139, 140, 141, 142, 143
What if I start the list with an even number? Then I have
(1) even 
(2) odd
(3) even
(4) odd
(5) even
(6) odd
(7) even The sum is
I can't get a whole number for n. I get 68 with 7 left over, so
I only have my 1st list.
My list of primes tell me 137 and 139 are prime.
The answer is 2.
Answer by dayne514(7)  (Show Source):
You can put this solution on YOUR website! Cosecutive integer are integers on a number line that follow eachother(ex; 3,4,and 5 are consecutive integers,
so let x = first consecutive integer
then x+1= second
x+2= third
x+3= fourth
x+4= fifth
x+5= sixth
x+6= seventh (since u are finding 7, represent each with a "let statement")
The problem says that the sum (addition)of these 7 consec integers is 980
so add all the expressions together and equal it to 980
x+x+1+x+2+x+3+x+4+x+5+x+6 = 980
now combine like terms
7x+21=980 now solve the equation
-21 -21
7x=959
divide by 7 on each side
x=137 now by substitution (use your let statements)find the other 6 numbers
x+1=138
x+2=139
x+3=140
X+4=141
x+5=142
x+6=143
the seven integers are 137,138,139,140,141,142,143, which one only have 1 and itself as factors? 137 and 139
Answer by jgr45(31)  (Show Source):
You can put this solution on YOUR website! Since we are asked to find an odd number of integers (in this case 7) we can use the method I've found to be a snap - let x = the MIDDLE number. The problem does not say odd/even, so therefore the difference between each number is 1. The sum is 980. So -
first number = x-3
second number = x-2
third number = x-1
fourth number = x
fifth number = x+1
sixth number = x+2
seventh number = x+3
(x-3)+(x-2)+(x-1)+x+(x+1)+(x+2)+(x+3) = 980
The associative property of addition says we can place (and remove) sets of parentheses anywhere we like, and the commutative property says we can add them in any order, without affecting the sum in either case. Now, watch what happens:
(x+x+x+x+x+x+x)+(-1+1)+(-2+2)+(-3+3) = 980
7x = 980 (WHAT??? Yes, the 1's, 2's and 3's cancel out!) :)
x = 140 (the middle number, remember)
The numbers are 137, 138, 139, 140, 141, 142, and 143.
Whenever you have an odd number of consecutive numbers, setting your variable to the middle number will result in a simple multiplication of your variable multiplied by the number of consecutive integers equaling the total.
As for the question of how many are prime, first, we can eliminate all the even numbers since those are all divisible by 2. If you know your rules of divisibility you can see that 143 is divisible by 11 (11x13=143), and 141 is divisible by 3 (3x47=141). That leaves 139 and 137, which are prime, so there are two primes in the set.
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