# Math Help - Alternative solution for problem involving consecutive numbers.

1. ## Alternative solution for problem involving consecutive numbers.

Problem: What numbers from 30 to 70 are the sum of two or more consecutive numbers?

I was able to answer it by simply trying out every number and testing it, and I got:

30 is the sum of all consecutive numbers from 4 to 8.
31 is the sum of all consecutive numbers from 15 to 16.
33 is the sum of all consecutive numbers from 3 to 8.
34 is the sum of all consecutive numbers from 7 to 10.
35 is the sum of all consecutive numbers from 2 to 8.
36 is the sum of all consecutive numbers from 1 to 8.
37 is the sum of all consecutive numbers from 18 to 19.
38 is the sum of all consecutive numbers from 8 to 11.
39 is the sum of all consecutive numbers from 4 to 9.
40 is the sum of all consecutive numbers from 6 to 10.
41 is the sum of all consecutive numbers from 20 to 21.
42 is the sum of all consecutive numbers from 3 to 9.
43 is the sum of all consecutive numbers from 21 to 22.
44 is the sum of all consecutive numbers from 2 to 9.
45 is the sum of all consecutive numbers from 1 to 9.
46 is the sum of all consecutive numbers from 10 to 13.
47 is the sum of all consecutive numbers from 23 to 24.
48 is the sum of all consecutive numbers from 15 to 17.
49 is the sum of all consecutive numbers from 4 to 10.
50 is the sum of all consecutive numbers from 8 to 12.
51 is the sum of all consecutive numbers from 6 to 11.
52 is the sum of all consecutive numbers from 3 to 10.
53 is the sum of all consecutive numbers from 26 to 27.
54 is the sum of all consecutive numbers from 2 to 10.
55 is the sum of all consecutive numbers from 1 to 10.
56 is the sum of all consecutive numbers from 5 to 11.
57 is the sum of all consecutive numbers from 7 to 12.
58 is the sum of all consecutive numbers from 13 to 16.
59 is the sum of all consecutive numbers from 29 to 30.
60 is the sum of all consecutive numbers from 4 to 11.
61 is the sum of all consecutive numbers from 30 to 31.
62 is the sum of all consecutive numbers from 14 to 17.
63 is the sum of all consecutive numbers from 3 to 11.
65 is the sum of all consecutive numbers from 2 to 11.
66 is the sum of all consecutive numbers from 1 to 11.
67 is the sum of all consecutive numbers from 33 to 34.
68 is the sum of all consecutive numbers from 5 to 12.
69 is the sum of all consecutive numbers from 9 to 14.
70 is the sum of all consecutive numbers from 7 to 13.

Since this is very tedious, I was wondering if there was any alternative solution for this.

Thanks.

2. ## Re: Alternative solution for problem involving consecutive numbers.

note ...

$\sum_{k=1}^n k = 1+2+3+ ... +n = \frac{n(n+1)}{2}$

3. ## Re: Alternative solution for problem involving consecutive numbers.

Hello, miguel11795!

What numbers from 30 to 70 are the sum of two or more consecutive numbers?

Every odd integer is the sum of two consecutive integers.

Given an odd $n$, subtract one and divide by two.
. . This gives us the first of the two integers.

Example: $n = 59 \quad\Rightarrow\quad \frac{59-1}{2} \,=\,29$
Therefore: . $59 \:=\:29 + 30$

Now consider the even numbers from 30 to 70.

Every multiple of 3 is the sum of three consecutive integers.

If $n$ is a multiple of 3, divide by 3.
. . This gives us the middle number.

Example: $n = 42 \quad\Rightarrow\quad \frac{42}{3} \,=\,14$
Therefore: . $42 \:=\:13 + 14 + 15$

Every multiple of 5 is the sum of five consecutive integers.

If $n$ is a multiple of 5, divide by 5.
. . This gives us the middle number.

Example: $n = 50 \quad\Rightarrow\quad \frac{50}{5} \,=\,10$
Therefore: . $50 \:=\:8 + 9 + 10 + 11 + 12$

Every multiple of 7 is the sum of seven consecutive integers.

If $n$ is a multiple of 7, divide by 7.
. . This gives us the middle number.

Example: . $n = 56 \quad\Rightarrow\quad \frac{56}{7} \,=\,8$
Therefore: . $56 \:=\:5 + 6 + 7 + 8 + 9 + 10 + 11$

$\text{This leaves us with: }\:32, 34, 38, 44, 46, 52, 58, 62, 64, 68$

Every odd multiple of 4 is the sum of eight consecutive integers.

If $n$ is an odd multiple of 4, divide by 4.
. . This gives us the sum of the middle two numbers.

Example: . $n = 52 \quad\Rightarrow\quad \frac{52}{4} = 13 \quad\Rightarrow\quad 13 \:=\:6+7$
Therefore: . $52 \:=\:3+4+5+6+7+8+9+10$

$\text{This leaves us with: }\:32,34,38,46,58,62,64$

Every number of the form $4k+2$ is the sum of four consecutive integers.
. . ( $n$ is two more than a multiple of 4,)

Subtract 6 and divide by 4.
. . This gives us the first of the four integers.

Example: . $n = 46 \quad\Rightarrow\quad \frac{46-6}{4} \,=\,10$
Therefore: . $46 \:=\:10+11+12 +13$

We are left with 32 and 64 (powers of 2)
. . which cannot be expressed as the sum of consecutive integers.