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06 Jun 2026, 12:10
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The average of consecutive integers is always half an integer. How many ways can 885 be written as the sum of at least two consecutive positive integers in ascending order?
(A) 7
(B) 8
(C) 14
(D) 15
(E) 16
(A) 7
(B) 8
(C) 14
(D) 15
(E) 16
06 Jun 2026, 22:25
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A number can be written as a sum of consecutive positive integers in as many ways as it has odd divisors,
excluding the one-term representation.
Factor 885:
[885 = 3 X 5 X 59]
Since all prime factors are odd, the number of odd divisors is:
[(1+1)(1+1)(1+1)=8]
So 885 can be written as a sum of consecutive integers in 8 ways.
One of these is the trivial representation:
[885]
But the problem asks for at least two consecutive positive integers, so exclude that one.
[8-1=7]
Answer: (A) 7.
excluding the one-term representation.
Factor 885:
[885 = 3 X 5 X 59]
Since all prime factors are odd, the number of odd divisors is:
[(1+1)(1+1)(1+1)=8]
So 885 can be written as a sum of consecutive integers in 8 ways.
One of these is the trivial representation:
[885]
But the problem asks for at least two consecutive positive integers, so exclude that one.
[8-1=7]
Answer: (A) 7.
06 Jun 2026, 23:02
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Answer is right, but the explanation is not.
This is actually a tricky problem.
You missed that 2 is a possibility. In an AP spread, when you have even number of consecutive integers, then the median = average = half of the middle two integers.
In other words 885 = 442+443 is a possibility.
2 is not a factor of 885 and is still included.
Why the odd divisors actually work?
Because again in an AP spread of consecutive integers, we have the middle value to be the integer itself. And the only we spread them like that is when the number itself is divisible by another integer (for all odd divisors).
Now,
885 = 3*5*59.
885/3 = 295, valid.
885/5 = 177, valid.
BUT
885/59 = 15.
Can we have a spread of positive integers with 15 as the middle value and having 58 more values? NO.
So 59 is NOT valid.
The integers that helps us give that spread are: 2,3,5 only.
We have 2 options for each.
2*2*2 = 8.
We have also included the case having 2*3*5 = 30.
Checking again:
885/30 = 29.5
So the 15th and 16th values will be:
29 and 30, valid again.
We have to however subtract the case with 1 as the divisor as 885/1 = 885 and we cant have 885 as a sum of 885 positive integers.
Thus 8-1 = 7.
Hope it helps.
======================================
It so happens that somehow having 59 in the set didn't matter because 2 substitutes it and we still land up with the same answer.
This is actually a tricky problem.
You missed that 2 is a possibility. In an AP spread, when you have even number of consecutive integers, then the median = average = half of the middle two integers.
In other words 885 = 442+443 is a possibility.
2 is not a factor of 885 and is still included.
Why the odd divisors actually work?
Because again in an AP spread of consecutive integers, we have the middle value to be the integer itself. And the only we spread them like that is when the number itself is divisible by another integer (for all odd divisors).
Now,
885 = 3*5*59.
885/3 = 295, valid.
885/5 = 177, valid.
BUT
885/59 = 15.
Can we have a spread of positive integers with 15 as the middle value and having 58 more values? NO.
So 59 is NOT valid.
The integers that helps us give that spread are: 2,3,5 only.
We have 2 options for each.
2*2*2 = 8.
We have also included the case having 2*3*5 = 30.
Checking again:
885/30 = 29.5
So the 15th and 16th values will be:
29 and 30, valid again.
We have to however subtract the case with 1 as the divisor as 885/1 = 885 and we cant have 885 as a sum of 885 positive integers.
Thus 8-1 = 7.
Hope it helps.
======================================
It so happens that somehow having 59 in the set didn't matter because 2 substitutes it and we still land up with the same answer.
AdarshSambare
