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# If the sum of all the even numbers from 1 to n(n: odd number) is 69*70

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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18 Jul 2017, 01:11
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45% (medium)

Question Stats:

71% (02:14) correct 29% (02:02) wrong based on 62 sessions

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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply Senior SC Moderator Joined: 22 May 2016 Posts: 3574 If the sum of all the even numbers from 1 to n(n: odd number) is 69*70 [#permalink] ### Show Tags 18 Jul 2017, 17:17 6 MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n? A. 135 B. 137 C. 139 D. 141 E. 143 1. If you don't remember the formula for summing consecutive even integers, you can use the general formula for the sum of any evenly spaced set where Sum = average (arithmetic mean) * number of terms The average (arithmetic mean) of evenly spaced consecutive numbers is First + Last divided by 2. 2. Given 69*70, from the basic formula you know that: -- one of those numbers is the average of the first and last terms, and -- the other number is the number of terms The first even number from 1 to n (where n is odd) is 2. Looking again at 69*70 . . . the average of two even terms (first and last) cannot be odd. So 69 must be the number of terms. And 70 must be the average of 2 and the last even term. 3. $$\frac{2 + Last Even}{2}$$= 70 2 + last even term = 140 Last even term = 138. n is odd. For 138 to be included in 1 to n . . . n must be 139. (Or, from the outset it should be clear that the last even term is n - 1.) Answer C _________________ SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here. Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver ##### General Discussion Director Joined: 04 Dec 2015 Posts: 743 Location: India Concentration: Technology, Strategy WE: Information Technology (Consulting) If the sum of all the even numbers from 1 to n(n: odd number) is 69*70 [#permalink] ### Show Tags 18 Jul 2017, 01:38 1 MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n? A. 135 B. 137 C. 139 D. 141 E. 143 Sum of first $$n$$ positive even numbers is given by formula $$= n(n+1)$$ Sum of all even numbers from $$1$$ to $$n = 69 * 70$$ Therefore $$n(n+1) = 69* 70$$ Even numbers from $$1$$ to $$n = 2, 4, 6, 8 , 10 .....$$ $$2$$ is the first even number. $$n = 69 * 2 + 1 = 139$$ Answer (C)... _________________ Please Press "+1 Kudos" to appreciate. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8027 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70 [#permalink] ### Show Tags 20 Jul 2017, 01:06 1 ==> If n=2m+1, the even numbers until 2m+1 becomes until 2m, and the sum of the even numbers is 2+4+…+2m=2(1+2+…m)=2m(m+1)/2=m(m+1). From m(m+1)=69*70, you get m=69, which becomes 2m+1=2(69)+1=139. The answer is C. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Joined: 12 Sep 2017
Posts: 302
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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02 Feb 2019, 15:17
sashiim20 wrote:
MathRevolution wrote:
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

Sum of first $$n$$ positive even numbers is given by formula $$= n(n+1)$$

Sum of all even numbers from $$1$$ to $$n = 69 * 70$$

Therefore $$n(n+1) = 69* 70$$

Even numbers from $$1$$ to $$n = 2, 4, 6, 8 , 10 .....$$

$$2$$ is the first even number.

$$n = 69 * 2 + 1 = 139$$

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Hello sashiim20 !

This was also my approach but I have a doubt, isn't n = 69?

n(n+1) = sum of first evens = 69(69+1)

Thank you so much in advance!
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Posts: 240
Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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06 Feb 2019, 19:17
MathRevolution wrote:
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

My reasoning:

Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end.

$$\frac{n}{2} * (2*2 + 2*(n-1)) = 69*70$$

Solve for n and we get n = 69.

This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C)
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Joined: 12 Sep 2017
Posts: 302
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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06 Feb 2019, 19:45
kchen1994 wrote:
MathRevolution wrote:
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

My reasoning:

Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end.

$$\frac{n}{2} * (2*2 + 2*(n-1)) = 69*70$$

Solve for n and we get n = 69.

This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C)

Hello kchen1994 !

How did you solve for n?

$$\frac{n}{2} * (2*2 + 2*(n-1)) = 69*70$$

Solve for n and we get n = 69.

I don't know what to do after it turns to be:

(2)(69)(70) = 2n + 2n2 = 2n(1+n)

(69)(70) = n(1+n)

How do you know that 69 is n?

Kind regards!
Senior Manager
Joined: 12 Sep 2017
Posts: 302
Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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06 Feb 2019, 19:58
MathRevolution wrote:
==> If n=2m+1, the even numbers until 2m+1 becomes until 2m, and the sum of the even numbers is 2+4+…+2m=2(1+2+…m)=2m(m+1)/2=m(m+1). From m(m+1)=69*70, you get m=69, which becomes 2m+1=2(69)+1=139.

Hello MathRevolution !

How do we know that n=2m+1?

Kind regards!
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Posts: 994
Location: India
Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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06 Feb 2019, 20:03
MathRevolution wrote:
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

So yes, what did i do here.

We have the sum as 69 * 70, this means in the formula for $$S_n$$ = n/2 (FT + LT)

n could have been divisible by 2 or FL + LT could have been divisible by 2

The only condition satisfying here was the former

69*2= 138

Key word : n: odd number

it will be the next term 139

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Joined: 22 Sep 2018
Posts: 240
Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70  [#permalink]

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06 Feb 2019, 20:12
jfranciscocuencag wrote:
kchen1994 wrote:
MathRevolution wrote:
If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?

A. 135
B. 137
C. 139
D. 141
E. 143

My reasoning:

Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end.

$$\frac{n}{2} * (2*2 + 2*(n-1)) = 69*70$$

Solve for n and we get n = 69.

This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C)

Hello kchen1994 !

How did you solve for n?

$$\frac{n}{2} * (2*2 + 2*(n-1)) = 69*70$$

Solve for n and we get n = 69.

I don't know what to do after it turns to be:

(2)(69)(70) = 2n + 2n2 = 2n(1+n)

(69)(70) = n(1+n)

How do you know that 69 is n?

Kind regards!

So if you multiply n in we get $$69(70) = n + n^2$$

Since 69*70 has already been factored for you you can solve this quadratic equation as so: $$(n+70)(n-69) = 0$$

A series can never be negative so n must be 69. Keep in mind this only accounts for all the even values in our series. We must include all the odd values so 69*2. But since we skipped 1 when we started our series, we need to add it back in. Does this make sense?
Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70   [#permalink] 06 Feb 2019, 20:12
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