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Re: DS: Sum of consecutive numbers [#permalink]
04 Apr 2008, 19:48

Question Tells us sum is of consecutive integers.

Statement 1: Tells us number of element is even. So it can 22+23 (2 even) = 45 or 1+ 2 + 3+ ...+ 9 (9 odd)= 45 or 5+6+7+...+10 (6 even) = 45, So insufficient.

Statement 2: Tells us total number of terms is less than 9. In all the cases considered above n is less than 9. So this statement alone is also insufficient.

Combining them together is not going to give any results as we are using same set of example for both the statements.

Re: DS: Sum of consecutive numbers [#permalink]
05 Apr 2008, 05:43

The formula n(n+1)/2 is only valid if you have series starting from 1. So it will be valid for 1+2+3+...+9 But it will not be valid for 4+5+6+7 or for that matter any series that does not start with 1.

Re: DS: sum of n consecutive ints [#permalink]
25 May 2008, 21:07

Expert's post

E

Thanks for a good problem. +1

sum=45. So, construct all sequences: n is odd. For odd n we can say that ((n+1)/2)th number is median and equal sum/n. In our case, 45 is divisible by odd n only if n is {3,5,9,15,45} (odd factors of 45) n=3: 45/3=15. 14,15,16 n=5: 45/5=9. 7,8,9,10,11 n=9: 45/5=5. 1,2,3,4,5,6,7,8,9 n=15,45: sequences will be include negative numbers that contradicts the problem.

n is even. For even n we can say that sum/n is median and has to be equidistant between (n/2)th and (n/2+1)th numbers. In other words, sum has to be decimal number like xxx,5

In our case, 45 is divisible by even n and result xxx.5 only if n is {2,6,10,18,30,90} n=2: 45/2=22.5. 22,(22.5),23 n=6: 45/6=7.5. 5,6,7,(7.5),8,9,10 n=10,18,30,90: sequences will be include negative numbers that contradicts the problem.

Re: DS: sum of n consecutive ints [#permalink]
26 May 2008, 00:31

Expert's post

alpha_plus_gamma wrote:

The sum of arthimetic progression (here with diff = 1) needs is

S = n(a1 + aN)/2 OR S = N [2*a1 + (N-1)d] /2

Thus if you know sum, you still have 2 unknowns,

(1) & (2) individually or together do not give any definite value of N,

So E.

But how can you use your approach for slightly modified problem: "The sum of n consecutive positive integers is 47. What is the value of n?" _________________

Re: DS: sum of n consecutive ints [#permalink]
26 May 2008, 01:50

Expert's post

Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it? _________________

Re: DS: sum of n consecutive ints [#permalink]
26 May 2008, 01:59

walker wrote:

Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it?

- I wouldn't say E is chosen bcoz of too many unknowns; rather given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E. - You are right; substitute 45 for 47 is insensitive to the substitution. - On a side note even the method you gave for counting 5s in 4500! question is an offshoot of this method.

Re: DS: sum of n consecutive ints [#permalink]
26 May 2008, 02:02

When dealing with the sum of evenly spaced integers (also known as arithmetic progression or sequence), just remember the simple, universal formula: Sum = Average* (# of values). [* represents the multiplication sign.] The space is 1 here. Applied to this problem, Sum = n*(a1+aN)/2. a1 and aN are the first term and the last term of the sequence respectively, (a1+ aN)/2 being the average, but also the median or the midpoint. They all coincide when dealing with an arithmetic progression. n is of course the number of terms. So, it follows that n(a1 + aN)/2 = 45 and (a1 + aN) = 90/n. Since a1 + a2 is an integer, so must be 90/n. It then boils down to finding the positive factors of 90. Let's find the prime factorization of 90. 90 = 9*10 = 3*3*2*5 = 3^2 * 2 * 5. We can easily list all the positive factors of 90; there are 12 of them. I actually have a formula to find them fast but that's another topic. 1/Since n is even, let's focus on the even values of n; they are 2, 6, 10, 18, 30, and 90. But if n = 10 then the median of the sequence will be (a1 + a2)/2 = 45/10 = 4.5 and some of the terms will be negative (NOT Allowed). Any n > 10 will have to be rejected for the same reason. Therefore n must be 2 or 6: UNSUFF. 2/ n < 9 includes more than one n (2, 3, 5, 6): UNSUFF. Together, we are still stuck with 2 and 6. Note that when n = 2, the average is 45/2 = 22.5; then a1 = 22 and aN = a2 = 23. 22+23 = 45 When n = 6, then a1 = 5; a2 = 6; a3 =7; a4 = 8, a5 = 9, a6 = 10. 5+6+7 +8+9+10 = 45. Regards,

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Last edited by TheGMATDoctor on 13 Jul 2010, 12:10, edited 2 times in total.

Re: DS: sum of n consecutive ints [#permalink]
26 May 2008, 02:03

farend wrote:

walker wrote:

Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it?

- I wouldn't say E is chosen bcoz of too many unknowns; rather given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E. - You are right; substitute 45 for 47 is insensitive to the substitution. - On a side note even the method you gave for counting 5s in 4500! question is an offshoot of this method.

I too meant >> given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E.