Sums of Consecutive Integers
In what ways can 1,000,000 be expressed as the sum of consecutive positive
whole numbers?
Source: Durango
Herald Puzzle Corner, by Dick Gibbs
Solution
The answer is 7 ways:
The 640 numbers 1243+1244+...+1882=1,000,000,
the 625 numbers 1288+1289+...+1912=1,000,000
the 128 numbers 7749+7750+...+7876=1,000,000
the 125 numbers 7938+7939+...+8062=1,000,000
the 25 numbers 39,988+39,989+...+40,012=1,000,000
the 5 numbers 199,998+199,999+...+200,002=1,000,000
and the 1 number 1,000,000.
Some people have raised a concern that there is no such thing as one
consecutive number. That is, the single number 1,000,000 is not itself an
expression of the sum of consecutive numbers. They point to the use of
plural in the question to say that at the very least, the question is deceptive
in its ambiguity. This is a fair concern, and perhaps the writers of the
puzzle intended this ambiguity as an extra degree of difficulty, frustrating as
that may be for some people!
An informal consensus of mathematicians is that a property, such as
"consecutive", is satisfied as long as no element of the sequence
violates that property. Thus, a sequence of one term has the property of
its terms being "consecutive" because there is no non-consecutive
term. For that matter, the sequence of zero terms can be said to be
consecutive for the same reason.
Ruth, a participant in the "Ask
NRICH" discussion group, pointed out that the question might also have
been posed,
In what ways can 1,000,000 be expressed as the difference of two triangular
numbers?
A more general problem
What math enthusiast can pass up the opportunity to pluck a general rule from
a specific instance? Here's a way to express a more general version of the
problem, the solution to which sheds light on this problem...
In what ways can N be expressed as the sum of one or more consecutive
positive integers?
If mn=N, with n odd, then n consecutive positive numbers have a "middle
number", which is the average of the n numbers. If the sum of the n
numbers is N, then this average is m.
Example: 3*7=21, so let m be 3 and n be the odd number of terms,
7. Then we can find 7 consecutive integers whose average is 3, and
therefore the sum of the 7 numbers is 21. These numbers are 0, 1, 2, 3,
4, 5, and 6. Oops! These aren't all positive integers, so
this isn't a solution to the question.
Now let 7*3=21, with m=7 the average of the terms, and n=3 is the number of
terms. Then we have 6+7+8=21, which is a solution to the question.
A little thought will convince you that an odd factor of N can be used to make
an odd number, n, of terms whose average is m and whose sum is N only if 2m
> n. This is true because the smallest term is m-n/2+1/2, which must
be greater than 1/2.
Let me use this thinking to refine my statement about sums of an odd number
of consecutive terms:
| If mn=N, with n odd, and 2m > n, then the n consecutive positive
numbers ranging from m-n/2+1/2 to m+n/2-1/2, has an average of m, and a
sum of N. |
What about an even number of consecutive positive numbers? Can they add
up to N? Let's see. An even number of consecutive numbers has no
"middle" number. The middle, if you can call it that, of the
numbers is the average of the middle two. It's also the average of the
first and last. And it's the average of the second and
second-to-last. I'm sure you get the idea: the average of an even number
of consecutive integers is a number midway between two integers, which we can
call n/2, where n is an odd number.
Let 2m be this even number of consecutive integers whose average is
n/2. Then, N, the sum of the integers is given by N=(2m)(n/2)=mn.
Example: Just for fun, let's use the same example as before: 3*7=21,
letting m be 3 and n be 7. So we can have 2m (six) terms whose average
is n/2 (3 1/2) that add up to 21. They are 1+2+3+4+5+6=21.
Now, let 7*3=21, with m=7 (so there will be 2m=14 terms) and n=3 (so the
average of the terms is n/2=3/2). These terms are -5+-4+-3+-2+-1+0+1+2+3+4+5+6+7+8=21.
Oops! Once again, we have a sequence that doesn't satisfy the problem
because some of the terms are not positive integers. In order for all
the terms to be positive, the smallest term, which is n/2-m+1/2, must be
greater than 1/2, so n > 2m.
As before, I will refine this second statement about the sums of an even
number of consecutive terms:
|
If mn=N, with n odd, and n > 2m, then the 2m consecutive positive
numbers ranging from n/2-m+1/2 to n/2+m-1/2, has an average of n/2, and a sum of
N.
|
Together, the two statements in the boxes, above, give a solution for each
odd divisor, n, of N. Each solution is a unique set of consecutive
positive integers whose sum is N.
Are there any other sequences of consecutive positive integers that sum to
N? No, because every sequence either has an even number of terms, in which
case the number of terms is 2m for some number, m, and their average is n/2 for
some odd number, n, so their sum is mn; --OR-- it has an odd number of terms, in
which case the number of terms is some odd number, n, and their average is an
integer, m, so their sum is mn. In either of these cases, every sequence
summing to the product mn is identified by one of the boxed statements, above.
If N=2kM, where M is odd, then the odd divisors of n is the same
as divisors of M. The number of such divisors can be obtained from the
prime factorization of M, by taking the product of the numbers one greater than
each exponent.
Example: Let's use the boxed statements to find the 7 sequences whose
sum is 1,000,000. The odd factors of 1,000,000 are 1, 5, 25, 125, 625,
3125, and 15625.
When n=15625, m=64, n > 2m, so 2*64 numbers averaging 15625/2 sum to: 7749+7750+...+7876=1,000,000
When n=3125, m=320, n > 2m, so 2*320 numbers averaging 3125/2 sum to: 1243+1244+...+1882=1,000,000
When n=625, m=1600, 2m > n, so 625 numbers averaging 1600 sum to: 1288+1289+...+1912=1,000,000
When n=125, m=8000, 2m > n, so 125 numbers averaging 8000 sum to: 7938+7939+...+8062=1,000,000
When n=25, m=40,000, 2m > n, so 25 numbers averaging 40,000 sum to: 39,988+39,989+...+40,012=1,000,000
When n=5, m=200,000, 2m > n, so 5 numbers averaging 200,000 sum to: 199,998+199,999+...+200,002=1,000,000
When n=1, m=1,000,000, 2m > n, so 1 number averaging 1,000,000 sums to: 1,000,000=1,000,000
Would you like another example? What if N is odd? Let N=15, which
has four odd divisors:
Example: N=15
When n=15, m=1, n > 2m, so 2*1 numbers averaging 15/2 sum to: 7+8=15
When n=5, m=3, 2m > n, so 5 numbers averaging 3 sum to: 1+2+...+5=15
When n=3, m=5, 2m > n, so 3 numbers averaging 5 sum to: 4+5+6=15
When n=1, m=15, 2m > n, so 1 number averaging 15 sums to: 15=15
And one more example:
Example: N = 25 33 52 7 = 151,200,
which has (3+1)(2+1)(1+1) = 24 odd divisors:
When n=4725, m=32, n > 2m, so 2*32 numbers averaging 4725/2 sum to: 2331+2332+...+2394=151,200
When n=1575, m=96, n > 2m, so 2*96 numbers averaging 1575/2 sum to: 692+693+...+883=151,200
When n=945, m=160, n > 2m, so 2*160 numbers averaging 945/2 sum to: 313+314+...+632=151,200
When n=675, m=224, n > 2m, so 2*224 numbers averaging 675/2 sum to: 114+115+...+561=151,200
When n=525, m=288, 2m > n, so 525 numbers averaging 288 sum to: 26+27+...+550=151,200
When n=315, m=480, 2m > n, so 315 numbers averaging 480 sum to: 323+324+...+637=151,200
When n=225, m=672, 2m > n, so 225 numbers averaging 672 sum to: 560+561+...+784=151,200
When n=189, m=800, 2m > n, so 189 numbers averaging 800 sum to: 706+707+...+894=151,200
When n=175, m=864, 2m > n, so 175 numbers averaging 864 sum to: 777+778+...+951=151,200
When n=135, m=1120, 2m > n, so 135 numbers averaging 1120 sum to: 1053+1054+...+1187=151,200
When n=105, m=1440, 2m > n, so 105 numbers averaging 1440 sum to: 1388+1389+...+1492=151,200
When n=75, m=2016, 2m > n, so 75 numbers averaging 2016 sum to: 1979+1980+...+2053=151,200
When n=63, m=2400, 2m > n, so 63 numbers averaging 2400 sum to: 2369+2370+...+2431=151,200
When n=45, m=3360, 2m > n, so 45 numbers averaging 3360 sum to: 3338+3339+...+3382=151,200
When n=35, m=4320, 2m > n, so 35 numbers averaging 4320 sum to: 4303+4304+...+4337=151,200
When n=27, m=5600, 2m > n, so 27 numbers averaging 5600 sum to: 5587+5588+...+5613=151,200
When n=25, m=6048, 2m > n, so 25 numbers averaging 6048 sum to: 6036+6037+...+6060=151,200
When n=21, m=7200, 2m > n, so 21 numbers averaging 7200 sum to: 7190+7191+...+7210=151,200
When n=15, m=10,080, 2m > n, so 15 numbers averaging 10,080 sum to:
10,073+10,074+...+10,087=151,200
When n=9, m=16,800, 2m > n, so 9 numbers averaging 16,800 sum to:
16,796+16,797+...+16,804=151,200
When n=7, m=21,600, 2m > n, so 7 numbers averaging 21,600 sum to:
21,597+21,598+...+21,603=151,200
When n=5, m=30,240, 2m > n, so 5 numbers averaging 30,240 sum to:
30,238+30,239+...+30,242=151,200
When n=3, m=50,400, 2m > n, so 3 numbers averaging 50,400 sum to:
50,399+50,400+50,401=151,200
When n=1, m=151,200, 2m > n, so 1 number averaging 151,200 sums to:
151,200=151,200
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