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