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.

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , . Bookmark the permalink.

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