In how many ways can 345 be written as the sum of an increasing sequen

Question

In how many ways can 345 be written as the sum of an increasing sequence of two or more consecutive positive integers?

(A) 1

(B) 3

(C) 5

(D) 6

(E) 7

CareerGeek · Answer

345/3 = 115 ---> {114, 115, 116}
345/5 = 69 ---> {67, 68, 69, 70, 71}

similarly look for other divisors of 345 = 1, 3, 5, 23, 15 (3x5), 69 (3x23), 115 (5x23), 345

345/15 = 23 ---> {16, 17 . . . . ,23, 24, 25, . . . . .30}
345/23 = 15 ---> {4, 5, 6, . . . . .15, 16, 17, . . . . . 25, 26}
345/69 = 5 ---> {-29, -27, - 26, . . . . . . 5, 6, 7, . . . . . . 39}
345/115 = 3 ---> {-54, -53, -52, . . . . 3, 4, 5, . . . . . . . 60}
345/345 = 1 ---> {-171, -170, -169, . . . . .,1, 2, 3, . . . . . . . 173}
345/1 = 345 ---> {345} -  Not Possible as question is asking atleast 2 positive consecutive integers!

So, total possible ways = 7
Option E

rupam29 · Answer

Hi,
Hope that the following might be helpful.
Suppose, there are n such consecutive positive numbers: a, a+1,a+2 .... a+n-1. 
So, we can write in the following way:
a+ (a+1)+(a+2)+ ....+( a+n-1)= 345
or, n*a + (1+2+.....+ n-1)= 345
or, n*a+ n*(n-1)/2 = 345
or, 2n*a+ n*(n-1)= 690
or, 2*a + (n-1)= 690/n
As both components of the left side of the equation are integers, the right part will be an integer too; that means that the number ‘n’ must be a divisor of 690. 
Now, the prime divisors of 690 are 2,3,5 and 23; so ‘n’ will be any one of these prime divisors or any product of them in any combination. 
If we proceed further, we will see that for seven such cases, i.e., for n=2,3,5,6,10,15 and 23, ‘a’ remains positive. For any other divisor, which is eventually bigger than those above-mentioned seven divisors, the value of ‘a’ will be negative.
Therefore, it can be concluded that there are 7 favorable ways.

Fdambro294 · Answer

I couldn't recognize the pattern quick enough.....taking 5 minutes, definitely can answer the Question.

2 Consecutive Integers is easy: X + (X + 1)
2X + 1 = 345
2X = 344

Since 344 is Even, then yes, 345 can be written as 2 Consecutive Integers

Then for Each Additional Consecutive Integer you test

1st)you make the Co-Efficient in front of X = (No. of Consec. Integers) * X

2nd) keeping adding an additional (X +1 to the Prior Term)

The Pattern will go like this

2 Consecutive Int. = 2x + 1 (1/2 of the Co-Efficient)

3 Consecutive Int. = 3x + 3 (1 Extra Number = Co-Efficient Added)

4 Consecutive Int. = 4x + 4 + 2 (1 Extra Number = Co-Efficient Added + 1/2 a Coefficient)

5 Consecutive Integers = 5x + 5 + 5 (2 Extra Numbers = Co-Efficient Added)

6 Consecutive Integers = 6X + 6 + 6 + 3 (2 Extra Numbers = Co-Efficient Added + 1/2 a Coefficient)

7 Consecutive Integers = 7X + 7 + 7 + 7 (3 Extra Numbers = Co-Efficient Added)

8 Consecutive Integers = 8X + 8 + 8 + 8 + 4 (3 Extra Numbers = Co-Efficient Added + 1/2 a Coefficient)

and the pattern will continue

I found 2 , 3 , 5 , 6 , 10 , and 15 consecutive integers worked. Then I saw the time lol

EDIT: the Post Above mine seems like a much more efficient use of time.

In how many ways can 345 be written as the sum of an increasing sequen

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In how many ways can 345 be written as the sum of an increasing sequen

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