## Sum of Consecutive Numbers

How many positive integers from 1 to 100 inclusive can be written as the sum of two or more consecutive positive integers?  (Different examples 10 and 51: 10 = 1 + 2 + 3 + 4 and 51 = 25 + 26)
Source: mathcontest.olemiss.edu 3/7/2011

SOLUTION
Sum of two consecutive numbers
$1+2=3$
$2+3=5$
$3+4=7$
$4+5=9$
$\cdots$
$48+49=97$
$49+50=99$

There are 100 numbers from 1 to 100. Half of them are even and half are odd. All the above sums are odd. Thus, there are 49 numbers from 3 to 99 that can be written as the sum of two consecutive positive integers.

Sum of three consecutive integers
$1+2+3=6$
$2+3+4=\underline{9}$        duplicates are indicated by underlines.
$3+4+5=12$
$4+5+6=\underline{15}$
$5+6+7=18$
$6+7+8=\underline{21}$
$\cdots$
$29+30+31=90$
$30+31+32=\underline{93}$
$31+32+33=96$
$32+33+34=\underline{99}$

An easy way to count these sums is to think of them as multiples of 6:
$6=6\times 1$
$12=6\times 2$
$18=6\times 3$
$\cdots$
$90=6\times 15$
$96=6\times 16$

Thus, sixteen numbers {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96} can be written as the sum of three consecutive positive integers.

Sum of four consecutive integers
﻿$1+2+3+4=10$
$2+3+4+5=14$
$3+4+5+6=\underline{18}$
$4+5+6+7=22$
$5+6+7+8=26$
$6+7+8+9=\underline{30}$
$7+8+9+10=34$
$8+9+10+11=38$
$9+10+11+12=\underline{42}$
$10+11+12+13=46$
$11+12+13+14=50$
$12+13+14+15=\underline{54}$
$13+14+15+16=58$
$14+15+16+17=62$
$15+16+17+18=\underline{66}$
$16+17+18+19=70$
$17+18+19+20=74$
$18+19+20+21=\underline{78}$
$19+20+21+22=82$
$20+21+22+23=86$
$21+22+23+24=\underline{90}$
$22+23+24+25=94$
$23+24+25+26=98$

Sixteen numbers {10, 14, 22, 26, 34, 38, 46, 50, 58, 62, 70, 74, 82, 86, 94, 98} can be written as the sum four consecutive positive integers.

Sum of five consecutive integers
$1+2+3+4+5=\underline{15}$
$2+3+4+5+6=20$
$3+4+5+6+7=\underline{25}$
$4+5+6+7+8=\underline{30}$        Surprise! 30 is a duplicate but 20 is not.
$5+6+7+8+9=\underline{35}$
$6+7+8+9+10=40$
$7+8+9+10+11=\underline{45}$
$8+9+10+11+12=\underline{50}$
$9+10+11+12+13=\underline{55}$
$10+11+12+13+14=\underline{60}$
$11+12+13+14+15=\underline{65}$
$12+13+14+15+16=\underline{70}$
$13+14+15+16+17=\underline{75}$
$14+15+16+17+18=80$
$15+16+17+18+19=\underline{85}$
$16+17+18+19+20=\underline{90}$
$17+18+19+20+21=\underline{95}$
$18+19+20+21+22=100$

Four numbers {20, 40, 80, 100} can be written as the sum of five consecutive positive integers.

Sum of six consecutive integers
$1+2+3+4+5+6=\underline{21}$
$2+3+4+5+6+7=\underline{27}$
$3+4+5+6+7+8=\underline{33}$
$4+5+6+7+8+9=\underline{39}$
$\cdots$

Luckily, all the sums are odd and therefore duplicates.

Sum of seven consecutive numbers
$1+2+3+4+5+6+7=28$
$2+3+4+5+6+7+8=\underline{35}$
$3+4+5+6+7+8+9=\underline{42}$
$4+5+6+7+8+9+10=\underline{49}$
$5+6+7+8+9+10+11=56$
$6+7+8+9+10+11+12=\underline{63}$
$7+8+9+10+11+12+13=\underline{70}$
$8+9+10+11+12+13+14=\underline{77}$
$9+10+11+12+13+14+15=\underline{84}$
$10+11+12+13+14+15+16=\underline{91}$
$11+12+13+14+15+16+17=\underline{98}$

Two numbers {28, 56} can be written as the sum of seven consecutive positive integers.

Sum of eight consecutive integers
$1+2+3+4+5+6+7+8=\underline{36}$
$2+3+4+5+6+7+8+9=44$
$3+4+5+6+7+8+9+10=52$
$4+5+6+7+8+9+10+11=\underline{60}$
$5+6+7+8+9+10+11+12=68$
$6+7+8+9+10+11+12+13=76$
$7+8+9+10+11+12+13+14=\underline{84}$
$8+9+10+11+12+13+14+15=92$
$9+10+11+12+13+14+15+16=\underline{100}$

Five numbers {44, 52, 68, 76, 92} can be written as the sum of eight consecutive positive integers.

Sum of nine consecutive integers
There are seven numbers {45, 54, 63, 72, 81, 90, 99} in this category but they are duplicates.

Sum of ten consecutive integers
There are six numbers {55, 65, 75, 85, 95} in this category but they are duplicates.

Sum of eleven consecutive integers
There are four numbers {66, 77, 88, 99} in this category; all are duplicates except 88. Thus, there is one number {88} that can be written as the sum of eleven consecutive positive integers.

Sum of twelve consecutive integers
There are two numbers {78, 90} in this category; both are duplicates.

Sum of thirteen consecutive integers
There is one number {91} in this category; it is a duplicate.

Sum of fourteen consecutive integers
$1+2+3+\cdots+13+14=105$.

The sum is greater than 100. So, we stop here.

Summary
Sum of two consecutive integers: 49 numbers
Sum of three consecutive integers: 16
Sum of four consecutive integers: 16
Sum of five consecutive integers: 4
Sum of six consecutive integers: 0
Sum of seven consecutive integers: 2
Sum of eight consecutive integers: 5
Sum of nine consecutive integers: 0
Sum of ten consecutive integers: 0
Sum of eleven consecutive integers: 1
Sum of twelve consecutive integers: 0
Sum of thirteen consecutive integers: 0

Total = $49+16+16+4+2+5+1=93$