I guess this a data sufficiency question and not PS
sum of n consecutive numbers starting with A as first number = ( (2A + n-1 ) / 2 ) n
( Sum of AP series ( ( 2a + (n-1)d )/ 2 )n here since its consecutive therefore d = 1 )
its given that the sum is 45
hence (2a + N - 1 )N = 90
factor of 90 = 2 * 5 * 9
hence n can be either 2 or 5 or 9
for n is odd we can have n = 5 or n = 9
for n >=9 we will have just n =9
so the second choice is sufficient.
Thanks!
-Mani
sum of consecutive terms
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hi.itz.mani
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rb90
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Is B the correct answer?because i am getting C as the answer.Here's how:
n(n+1)/2=45
= n^2 + n = 90
= n^2 + n - 90 = 0
= (n+10)(n-9) = 0
which gives us n=9 (since n cant be -ve)
Is this correct?
Please let me know
Thanks
n(n+1)/2=45
= n^2 + n = 90
= n^2 + n - 90 = 0
= (n+10)(n-9) = 0
which gives us n=9 (since n cant be -ve)
Is this correct?
Please let me know
Thanks
I think n(n+1)/2 is a formula used for the FIRST n nos. OA is definitely correct.rb90 wrote:Is B the correct answer?because i am getting C as the answer.Here's how:
n(n+1)/2=45
= n^2 + n = 90
= n^2 + n - 90 = 0
= (n+10)(n-9) = 0
which gives us n=9 (since n cant be -ve)
Is this correct?
Please let me know
Thanks
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Brian@VeritasPrep
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- Joined: Thu Jul 03, 2008 1:23 pm
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- GMAT Score:750
Hey everyone:
The OA is correct here and it's a pretty good question. Ultimately this one is more of a divisibility problem than anything else.
We need the sum of the digits to be 45, and that sum is going to be made of up of x + (x+1) + (x+2).... for n number of values. x in this case will be the first value, and each one goes up by one from there.
In the first statement, we know that we need an odd number of terms, so we could have:
x + x+1 + x+2 = 45
or
x + x+1 + x+2 + x+3 + x+4 = 45
The key on these is to determine whether, once we combine like terms, we'll get integer values of x in each case. For the first, we'd have:
3x + 3 = 45
3x = 42
because 42 is divisible by 3, we'll end up with an integer value of x, so we CAN have 3 consecutive integers.
For the second, we'd have:
5x + 10 = 45
5x = 35
because 35 is divisible by 5, we'll have an integer value of x, so we CAN have 5 consecutive integers and therefore statement 1 is not sufficient - we can have either 3 or 5 values.
For Statement 2, we know that the minimum number of values is 9, which would give us:
9x + (1+2+3+4+5+6+7+8) = 45
9x + 36 = 45
9x = 9
x = 1
So we can have 9 values...but can we have 10 or more?
For 10 values, the setup would be:
10x + 36 (from before) + 9 (for that last new integer) = 45
10x + 45 = 45
10x = 0
x = 0, which is not possible because we need POSITIVE integers. Therefore, the only number of values we can have is 9, because anything larger would give us either 0 or a negative number. Because statement 2 is sufficient, the correct answer is B.
The OA is correct here and it's a pretty good question. Ultimately this one is more of a divisibility problem than anything else.
We need the sum of the digits to be 45, and that sum is going to be made of up of x + (x+1) + (x+2).... for n number of values. x in this case will be the first value, and each one goes up by one from there.
In the first statement, we know that we need an odd number of terms, so we could have:
x + x+1 + x+2 = 45
or
x + x+1 + x+2 + x+3 + x+4 = 45
The key on these is to determine whether, once we combine like terms, we'll get integer values of x in each case. For the first, we'd have:
3x + 3 = 45
3x = 42
because 42 is divisible by 3, we'll end up with an integer value of x, so we CAN have 3 consecutive integers.
For the second, we'd have:
5x + 10 = 45
5x = 35
because 35 is divisible by 5, we'll have an integer value of x, so we CAN have 5 consecutive integers and therefore statement 1 is not sufficient - we can have either 3 or 5 values.
For Statement 2, we know that the minimum number of values is 9, which would give us:
9x + (1+2+3+4+5+6+7+8) = 45
9x + 36 = 45
9x = 9
x = 1
So we can have 9 values...but can we have 10 or more?
For 10 values, the setup would be:
10x + 36 (from before) + 9 (for that last new integer) = 45
10x + 45 = 45
10x = 0
x = 0, which is not possible because we need POSITIVE integers. Therefore, the only number of values we can have is 9, because anything larger would give us either 0 or a negative number. Because statement 2 is sufficient, the correct answer is B.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
Like always, wonderful explanation Brian!!!Brian@VeritasPrep wrote:Hey everyone:
The OA is correct here and it's a pretty good question. Ultimately this one is more of a divisibility problem than anything else.
We need the sum of the digits to be 45, and that sum is going to be made of up of x + (x+1) + (x+2).... for n number of values. x in this case will be the first value, and each one goes up by one from there.
In the first statement, we know that we need an odd number of terms, so we could have:
x + x+1 + x+2 = 45
or
x + x+1 + x+2 + x+3 + x+4 = 45
The key on these is to determine whether, once we combine like terms, we'll get integer values of x in each case. For the first, we'd have:
3x + 3 = 45
3x = 42
because 42 is divisible by 3, we'll end up with an integer value of x, so we CAN have 3 consecutive integers.
For the second, we'd have:
5x + 10 = 45
5x = 35
because 35 is divisible by 5, we'll have an integer value of x, so we CAN have 5 consecutive integers and therefore statement 1 is not sufficient - we can have either 3 or 5 values.
For Statement 2, we know that the minimum number of values is 9, which would give us:
9x + (1+2+3+4+5+6+7+8) = 45
9x + 36 = 45
9x = 9
x = 1
So we can have 9 values...but can we have 10 or more?
For 10 values, the setup would be:
10x + 36 (from before) + 9 (for that last new integer) = 45
10x + 45 = 45
10x = 0
x = 0, which is not possible because we need POSITIVE integers. Therefore, the only number of values we can have is 9, because anything larger would give us either 0 or a negative number. Because statement 2 is sufficient, the correct answer is B.
