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# How many different ways can you express the number 60 as a sum of consecutive integers?

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How many different ways can you express the number 60 as a sum of consecutive integers?

Jul 2, 2019

#2
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Let  n  be the value of the smallest addend. Let's find the sums of the first few consecutive integers starting at  n .

When the number of addends is  2 ,   the sum  =  2n + 1

When the number of addends is  3 ,   the sum  =  3n + 2 + 1

When the number of addends is  4 ,   the sum  =  4n + 3 + 2 + 1

When the number of addends is  5 ,   the sum  =  5n + 4 + 3 + 2 + 1

When the number of addends is  6 ,   the sum  =  6n + 5 + 4 + 3 + 2 + 1

When the number of addends is  7 ,   the sum  =  7n + 6 + 5 + 4 + 3 + 2 + 1  =  7n + 7 + 7 + 7  =  7n + 7(7 - 1)/2

When the number of addends is  a ,   the sum  =  an + a(a - 1) / 2

So we want to find the integer solutions to this equation:

an + a(a - 1) / 2   =   60

I don't know how to do that........so I had to use WolframAlpha to solve it. See:

https://www.wolframalpha.com/input/?i=integer+solutions:+a*n+%2B+a(a+-+1)+%2F+2%3D60,+a%3E0

Assuming  a > 1 ,  there are  7  different ways to express the number  60  as a sum of consecutive integers:

19 + 20 + 21  =  60

10 + 11 + 12 + 13 + 14  =  60

4 + 5 + 6 + 7 + 8 + 9 + 10 + 11  =  60

-3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11  =  60

-9 + -8 + -7 + -6 + . . . + 9 + 10 + 11 + 12 + 13 + 14  =  60

-18 + -17 + -16 + . . . + 18 + 19 + 20 + 21  =  60

-59 + -58 + -57 + . . . + 59 + 60  =  60

Jul 2, 2019

#1
+1   Jul 2, 2019
edited by CPhill  Jul 2, 2019
#2
+5

Let  n  be the value of the smallest addend. Let's find the sums of the first few consecutive integers starting at  n .

When the number of addends is  2 ,   the sum  =  2n + 1

When the number of addends is  3 ,   the sum  =  3n + 2 + 1

When the number of addends is  4 ,   the sum  =  4n + 3 + 2 + 1

When the number of addends is  5 ,   the sum  =  5n + 4 + 3 + 2 + 1

When the number of addends is  6 ,   the sum  =  6n + 5 + 4 + 3 + 2 + 1

When the number of addends is  7 ,   the sum  =  7n + 6 + 5 + 4 + 3 + 2 + 1  =  7n + 7 + 7 + 7  =  7n + 7(7 - 1)/2

When the number of addends is  a ,   the sum  =  an + a(a - 1) / 2

So we want to find the integer solutions to this equation:

an + a(a - 1) / 2   =   60

I don't know how to do that........so I had to use WolframAlpha to solve it. See:

https://www.wolframalpha.com/input/?i=integer+solutions:+a*n+%2B+a(a+-+1)+%2F+2%3D60,+a%3E0

Assuming  a > 1 ,  there are  7  different ways to express the number  60  as a sum of consecutive integers:

19 + 20 + 21  =  60

10 + 11 + 12 + 13 + 14  =  60

4 + 5 + 6 + 7 + 8 + 9 + 10 + 11  =  60

-3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11  =  60

-9 + -8 + -7 + -6 + . . . + 9 + 10 + 11 + 12 + 13 + 14  =  60

-18 + -17 + -16 + . . . + 18 + 19 + 20 + 21  =  60

-59 + -58 + -57 + . . . + 59 + 60  =  60

hectictar Jul 2, 2019
#3
+1

THX, hectictar   !!!   CPhill  Jul 2, 2019