45 caliber

[intrepid17]

Is there any other way to solve following problem , other than explained in CAT explanation ?:

The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9


Explanation given :

The possible values of n should be computed right away, to rephrase and simplify the question. Note that n consecutive positive integers that sum to 45 have a mean of 45/n, which is also the median of the set; therefore, the set must be arranged around 45/n. Also, any set of consecutive integers must have either an integer mean (if the number of integers is odd) or a mean that is an integer + 1/2 (if the number of integers is even). So, if we compute 45/n and see that it is neither an integer nor an integer +1/2, then we can eliminate this possibility right away.

Setting up a table that tracks not only the value of n but also the value of 45/n is useful.

n 45/n n positive consecutive integers summing to 45

1 45 45

2 22.5 22, 23

3 15 14, 15, 16

4 11.25 none

5 9 7, 8, 9, 10, 11

6 7.5 5, 6, 7, 8, 9, 10

7 6 3/7 none

8 5 5/8 none

9 5 1, 2, 3, 4, 5, 6, 7, 8, 9

10 4.5 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 -- but this doesn't work, because not all are positive integers

... ... impossible (the set will include negative integers, if an integer set can be found at all)



(1) INSUFFICIENT: If n is even, n could be either 2 or 6. Statement (1) is NOT sufficient.

Alternatively, to find these values algebraically, you can use the following procedure.
The sum of two consecutive integers can be represented as n + (n + 1) = 2n + 1
The sum of three consecutive integers = n + (n + 1) + (n + 2) = 3n + 3
The sum of four consecutive integers = 4n + 6
The sum of five consecutive integers = 5n + 10
The sum of six consecutive integers = 6n + 15

Since the expressions 2n + 1 and 6n + 15 can both yield 45 for integer values of n, 45 can be the sum of two or six consecutive integers.

(2) INSUFFICIENT: If n < 9, n could again take on either of the values 2 or 6 (or 3 or 5 according to the table or the expressions above)

(1) and (2) INSUFFICIENT: if we combine the two statements, n must be even and less than 9, so n could still be either of the values: 2 or 6.

The correct answer is E.

[tim]

Of course there is. But remember you only have 75 minutes to complete the math section, so the advanced number theory you learned in grad school is probably overkill. :) Notice how the statements give very vague information about the number; this is usually a giveaway that the best way to solve the problem is to rephrase the question to identify possible values of n..

[mailtobook]

Hi there,

The one thing I cant understand is how do you arrive at this conclusion with the algebraic approach:

"Since the expressions 2n + 1 and 6n + 15 can both yield 45 for integer values of n, 45 can be the sum of two or six consecutive integers"

Wouldnt "The sum of five consecutive integers = 5n + 10" also yield an integer for n when 45? 45-10=35/5=7=n

Im sure it doesnt work like that, but I cant figure out why from the explanation...

[jnelson0612]

"Since the expressions 2n + 1 and 6n + 15 can both yield 45 for integer values of n, 45 can be the sum of two or six consecutive integers"

Wouldnt "The sum of five consecutive integers = 5n + 10" also yield an integer for n when 45? 45-10=35/5=7=n

Im sure it doesnt work like that, but I cant figure out why from the explanation...

You're absolutely right. n could be 5, with numbers 7, 8, 9, 10, 11 summing to 45. However, we are only interested in even values for n for statement 1.

[atul.prasad]

Sum of n consecutive numbers:
= (n/2)(2a+n-1) = 45 , where a is the first number of the series

now from statement 1
n is even,

so n could be 2,4,6,8...
for n=2 , the numbers could be 22,23
you can leave out n=4 since it shall contain 2 odd and 2 even numbers, the sum of which can't be 45(odd)
for n=6 , the numbers would be 5,6,7,8,9,10 (just put n=6 and evaluate for a)

So Statement 1 is insufficient since we can't conclude on a single value of n

From statement 2
n < 9
we already inferred from statement 1 that for both n=2 and n=6 the equation holds true. So we can't conclude on a value for n
so statement 2 is also insufficient

Together: n is even and n is < 9
Same logic as for statement 2 can be used.

Answer should be E

[jnelson0612]

As always, correct atul.

[rajdeep115]

Sum of n consecutive numbers:
= (n/2)(2a+n-1) = 45 , where a is the first number of the series

now from statement 1
n is even,

so n could be 2,4,6,8...
for n=2 , the numbers could be 22,23
you can leave out n=4 since it shall contain 2 odd and 2 even numbers, the sum of which can't be 45(odd)
for n=6 , the numbers would be 5,6,7,8,9,10 (just put n=6 and evaluate for a)

So Statement 1 is insufficient since we can't conclude on a single value of n

From statement 2
n < 9
we already inferred from statement 1 that for both n=2 and n=6 the equation holds true. So we can't conclude on a value for n
so statement 2 is also insufficient

Together: n is even and n is < 9
Same logic as for statement 2 can be used.

Answer should be E

Great explanation. exactly what I was looking for...

[jnelson0612]

Great! Thanks a ton atul!

[roopesh.u]

I think there should be a d , the common difference between the terms in the equation.

Sum of n consecutive numbers:
= (n/2)[2a+(n-1)d] = 45

[tim]

i wouldn't worry either way. instead, focus on the fact that the sum of any evenly spaced set is the number of terms times the average and don't bother with such super-specific formulas..

[geezer0305]

I think there should be a d , the common difference between the terms in the equation.

Sum of n consecutive numbers:
= (n/2)[2a+(n-1)d] = 45

Yup, this is how i handled this situation :)

consecutive terms are in AP with a common difference of 1.

Thus, 45= (n/2)[2a+(n-1)*1]

we have one equation and two variables and both the statements in the question stem do not give us enough information to find the value of n.

[tim]

thanks for sharing. let us know if there are any further questions on this one..