## 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 five 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.

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

Answer: $93$

Alternative solution
Let $y=x+(x+1)+(x+2)+(x+3)+\cdots+(x+(n-3))+(x+(n-2))+(x+(n-1))$ be the sum of $n$ consecutive integers with the first term equal $x$. We apply the sum formula $y=(2x+n-1)n/2$ for various values of $n$.

$n=2$
$y=(2x+2-1)2/2=2x+1$
$2x+1$ is an odd integer. The $49$ integers are $\{3,5,7,\cdots,95,97,99\}$.

$n=3$
$y=(2x+3-1)3/2=3x+3$
$3\cdot 1+3=6$
$3\cdot 2+3=9$
$3\cdot 3+3=12$
$\cdots$
$3\cdot 30+3=93$
$3\cdot 31+3=96$
$3\cdot 32+3=99$
The $16$ odd integers are duplicates of case $n=2$. The remaining $16$ even integers are $\{6,12,18,\cdots,90,96\}$.

$n=4$
$y=(2x+4-1)4/2=4x+6$
$4\cdot 1+6=10$
$4\cdot 2+6=14$
$4\cdot 3+6=18$
$\cdots$
$4\cdot 21+6=90$
$4\cdot 22+6=94$
$4\cdot 23+6=98$
The $7$ multiples of $6$ are duplicates of case $n=3$. The remaining $16$ integers are $\{10,14,22,26,34,38,46,50,58,62,70,74,82,86,94,98\}$.

$n=5$
$y=(2x+5-1)5/2=5x+10$
$5\cdot 1+10=15$
$5\cdot 2+10=20$
$5\cdot 3+10=25$
$\cdots$
$5\cdot 16+10=90$
$5\cdot 17+10=95$
$5\cdot 18+10=100$
Fourteen of the integers are duplicates of case $n=3$ and $n=4$. The remaining $4$ integers are $\{20,40,80,100\}$.

$n=6$
$y=(2x+6-1)6/2=6x+15$
$6\cdot 1+15=21$
$6\cdot 2+15=27$
$6\cdot 3+15=33$
$\cdots$
$6\cdot 12+15=87$
$6\cdot 13+15=93$
$6\cdot 14+15=99$
All the $14$ integers are odd and duplicates of case $n=2$.

$n=7$
$y=(2x+7-1)7/2=7x+21$
$7\cdot 1+21=28$
$7\cdot 2+21=35$
$7\cdot 3+21=42$
$\cdots$
$7\cdot 9+21=84$
$7\cdot 10+21=91$
$7\cdot 11+21=98$
Nine of the integers are duplicates. The remaining $2$ integers are $\{28,56\}$.

$n=8$
$y=(2x+8-1)8/2=8x+28$
$8\cdot 1+28=36$
$8\cdot 2+28=44$
$8\cdot 3+28=52$
$\cdots$
$8\cdot 7+28=84$
$8\cdot 8+28=92$
$8\cdot 9+28=100$
Four of the integers are duplicates. The $5$ integers are $\{44,52,68,76,92\}$.

$n=9$
$y=(2x+9-1)9/2=9x+36$
$9\cdot 1+36=45$
$9\cdot 2+36=54$
$9\cdot 3+36=63$
$9\cdot 4+36=72$
$9\cdot 5+36=81$
$9\cdot 6+36=90$
$9\cdot 7+36=99$
All the $7$ integers are duplicates.

$n=10$
$y=(2x+10-1)10/2=10x+45$
$10\cdot 1+45=55$
$10\cdot 2+45=65$
$10\cdot 3+45=75$
$10\cdot 4+45=85$
$10\cdot 5+45=95$
All the $5$ integers are duplicates.

$n=11$
$y=(2x+11-1)11/2=11x+55$
$11\cdot 1+55=66$
$11\cdot 2+55=77$
$11\cdot 3+55=88$
$11\cdot 4+55=99$
All the integers are duplicates except $88$.

$n=12$
$y=(2x+12-1)12/2=12x+66$
$12\cdot 1+66=78$
$12\cdot 2+66=90$
Both integers are duplicates.

$n=13$
$y=(2x+13-1)13/2=13x+78$
$13\cdot 1+78=91$
$91$ is a duplicate.

$n=14$
$y=(2x+14-1)14/2=14x+91$
$14\cdot 1+91=105>100$ so we stop.