tricky one

[willbeatthegmat]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

[logitech]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

between 1 - 79

2+4+6+.....76+78

2( 1+2+3+...38+39)

2 x (39 x 40) / 2 = 39x40

so the answer you have is incorrect

[bluementor]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

Your answer (n = 159) is correct.

https://www.beatthegmat.com/difficult-ma ... 06-15.html

-BM-

[jdkster]

Sum of even numbers by definition cannot be odd, so 159 just can't be right

[iamcste]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

Simple e.g:

sum (2+4) =2*3 =n(n+1)

sum(2+4+6)=3*4=12=n(n+1)

Generalizing sum of n no. of even nos is n(n+1)

Apply this

sum ( even nos btn 1 and n)=79*80=79*(79+1)

hence, n=no of even terms =79


Hope this makes it clear

[Mom4MBA]

numbers are from 1-n
we need to take just the even numbers for sum
Given n is odd, so the last even number is n-1

Sum = 2+4+6+......+(n-1)
take 2 common
Sum = 2(1+2+3..........+(n-1)/2)
let (n-1)/2 = a
Sum = 2(1+2+3..........+a)
In the bracket we have the sum of a natural numbers from 1-a, use the standard formula for the sum
Sum = 2[a(a+1)/2]
Sum = a(a+1)

Given sum = 79*80

Sum = a(a+1) = 79*80
so a=79
substitute a=(n-1)/2

we get n=159, this is the correct answer

[sumanr84]

Easy but tricky

Number of even terms = 79, but the Q is asking for value of n ( that is 1....n ) = 159

[Mom4MBA]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

We need to add only the even numbers from 1 to n; when given n is odd that means that the last even number is (n-1)

So the sum will be of numbers 2,4,6,8,10,12,...............,(n-1), which are even

In all these are (n-1)/2 numbers that are to be added

[peddisetty]

Numbers are 2,4,6,8.......n. You can write them as 2(1,2,3.............n/2).

==>( 2*n/2*(n/2+1))/2 ==> n/2(n/2+1) = 79*80. n/2 = 79. n = 158. As n is odd, answer is 159. Your answer 159 is 100% correct.

[fluke]

Let me try to clear the confusion:

Between 1 and n(it is some odd number and not number of terms of the aritmetic progression); there are certain number of even numbers that adds up to 79*80

**********************************
For example:
Between 1, 11=n
Where 1 is an odd number and 11 is also an odd number:
x = total count of {2,4,6,8,10} = 5
there are total 5 even numbers sum of which is 5*6=30

So the number of terms
x=5 but n=11(it is value of unknown odd number that is one more than the last number of the AP 2,4,6,8,10)
************************************

Now we have to consider all even numbers between 1, n
We can say the even numbers will be 2, 4, 6, 8, 10, ......, n-1
Let 'x' be the number of terms of the arithmetic progression 2, 4, 6, 8, 10, ......, n-1

The sum of the above AP can be written as

S= x/2[2a+(x-1)d]
= x/2[2*2+(x-1)2]
= x[2+ (x-1]
= 2x + x^2 - x
= x^2 +x
= x * (x + 1)

Now we can equate:
x * (X+1) = 79 * 80

Thus x=79. So far we have just found the number of TERMS of the AP

if 79 is the number of terms. Then the last number of AP will be

Ax = a+(x-1)d = 2 + (79 - 1)2 = 2 + 78 *2 = 2 + 156 = 158

And we know from the question that the odd number "n" is one more than the last number of the arithmetic progression. Thus, the number is 159.

Hope this clarifies!!!

~fluke

[GMATGuruNY]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

The problem becomes less tricky if you realize that 79*80 represents the formula for determining the sum of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

Thus, 79 = the number of even integers and 80 = the average of the biggest and smallest.
Let x = biggest integer. Smallest integer = 2.
(x+2)/2 = 80.
x+2 = 160
x = 158.
Thus, 79*80 represents the sum of the even integers from 2 to 158.
Since n is the next biggest odd integer, n=159.

Below are more examples of determining the sum of even spaced integers. Again, here's the formula:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

From 1-10, the number of integers = (5-1)/1 + 1 = 5.
From 2-10, the number of even integers = (10-2)/2 + 1 = 5.
From 3 to 9, the number of multiples of 3 = (9-3)/3 + 1 = 3.

Thus:

The sum of the integers from 1 to 10 = 10 * (10+1)/2 = 55.
The sum of the even integers from 2 to 10 = 5 * (10+2)/2 = 30.
The sum of the multiples of 3 from 3 to 9 = 3 * (9+3)/2 = 18.

Hope this helps!

[Deepthi Subbu]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...[/quote]

The problem becomes less tricky if you realize that 79*80 represents the formula for determining the sum of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

Thus, 79 = the number of even integers and 80 = the average of the biggest and smallest.
Let x = biggest integer. Smallest integer = 2.

(x+2)/2 = 80.
x+2 = 160
x = 158.
Thus, 79*80 represents the sum of the even integers from 2 to 158.
Since n is the next biggest odd integer, n=159.

Below are more examples of determining the sum of even spaced integers. Again, here's the formula:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

From 1-10, the number of integers = (5-1)/1 + 1 = 5.
From 2-10, the number of even integers = (10-2)/2 + 1 = 5.
From 3 to 9, the number of multiples of 3 = (9-3)/3 + 1 = 3.

Thus:

The sum of the integers from 1 to 10 = 10 * (10+1)/2 = 55.
The sum of the even integers from 2 to 10 = 5 * (10+2)/2 = 30.
The sum of the multiples of 3 from 3 to 9 = 3 * (9+3)/2 = 18.

Hope this helps

[Deepthi Subbu]

Q-The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

The answer i have is 79 whereas i m gettin 159...

The problem becomes less tricky if you realize that 79*80 represents the formula for determining the sum of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

Thus, 79 = the number of even integers and 80 = the average of the biggest and smallest.
Let x = biggest integer. Smallest integer = 2.
(x+2)/2 = 80.
x+2 = 160
x = 158.
Thus, 79*80 represents the sum of the even integers from 2 to 158.
Since n is the next biggest odd integer, n=159.

Below are more examples of determining the sum of even spaced integers. Again, here's the formula:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

From 1-10, the number of integers = (5-1)/1 + 1 = 5.
From 2-10, the number of even integers = (10-2)/2 + 1 = 5.
From 3 to 9, the number of multiples of 3 = (9-3)/3 + 1 = 3.

Thus:

The sum of the integers from 1 to 10 = 10 * (10+1)/2 = 55.
The sum of the even integers from 2 to 10 = 5 * (10+2)/2 = 30.
The sum of the multiples of 3 from 3 to 9 = 3 * (9+3)/2 = 18.

Hope this helps!

Sum = 79*80
Sum = Average*no of terms
Average = (n-1+2)/2
=(n+1)/2
No of terms = (n-1+2)/2 +1
=(n+3)/2
79*80 = (n+1)/2 * (n+3)/2
When I substitute for n= 159, I get
25920 = 25280

Whats wrong?