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## Difficult Math Question #1

##### This topic has expert replies
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by k_abhi19 » Tue Sep 23, 2008 12:20 pm
n=159 should be the answer... I used a very simple approach.... by taking few cases...... for eg n=3........ in this case the sum of all even nos is 2.. or 1*2 .... similarly for n=5.. sum of all even nos between 1 & 5 will be 6 or 2*3 .. n=2+3=5.... for n=7... sum of all even nos in between will be 12 or 3*4 where 3+4=7=n.... thus when sum of all even nos bettwen 1 & n where n is odd... is 79*80.. then n=79+80 or 159....

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by davy420 » Tue Sep 23, 2008 1:49 pm
80 + 79 = 159

Can this problem be solved this way?

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### 79*80

by Xins » Fri Oct 31, 2008 10:56 am

I did scratch my head for a few minutes.

Heres what I found

Lets take n = 3 Sum of all even number from 1-3 = 2 = 2x1
Lets take n = 5 Sum of all even number from 1-3 = 6 = 2X3
Lets take n = 7 Sum of all even number from 1-3 = 12 = 4X3

I used the X pattern since the question stem says 79*80 i.e product of 2 consequtive numbers

All the above product needs to be represented in terms on n to get their n

2X1 = [ 2 -(2-1)/2] [ 2- (2+1)/2 ]
3x2 = [ 5 -(5-1)/2] [ 5- (5+1)/2 ]
4X2 = [ 7 -(7-1)/2] [ 7- (7+1)/2 ]

Hence 80 X 79 = [n- (n-1)/2] [n -(n+1)/2]

If you solve for n ; n = 159

Hope this helps

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### 79*80

by Xins » Fri Oct 31, 2008 10:59 am
Pardon the typo above. It should be 4X3

4X3 = [ 7 -(7-1)/2] [ 7- (7+1)/2 ]

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### 79*80

by Xins » Fri Oct 31, 2008 1:27 pm
I am on a roll of typos today ......

2X1 = [ 2 -(2-1)/2] [ 2- (2+1)/2 ] should be
2X1 = [ 3 -(3-1)/2] [ 3- (3+1)/2 ]

Sorry again.......But I hope you get the point.........

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by earth@work » Sun Nov 02, 2008 6:53 pm
nice question 800guy

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by piyushdabomb » Sun Nov 02, 2008 8:50 pm
What difficulty level is this sort of question? approximation of 650?
-------------------
Sincerely,

Piyush A.

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### Was Confused for a sec as well

by gmatbedamned » Tue Nov 18, 2008 4:31 pm
Love the list of tougher questions!!

I think we have a confusion of variables by the folks giving 79 as the answer.

I assume they are using the equation Sum(n terms)= n/2 * (2a + (n-1)d) where a is the first term (2) and d is the common difference (2). The equation yields "79" as the answer. Which is right, there are 79 TERMS.

The confusion stems from the fact the original question refers to n as the last term wihich is odd, NOT the number of terms. After you get the n=79 terms from the equation, to find the largest even term = #of terms * common differnece: 79*2 = 158. But n is odd and must be greater than 158 and less than 160. So n =159.

I personally like the equation Sum(X Terms) = X(First+Last)/2. The first is 2 and the last is n-1 and the number of terms (X) is (n-1)/2. It is a messier caculation but more fun!

Anyway, my two cents...

Sorry for reposting a dead topic but I was confused and I thought others would be helped by the clarification.

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by jeffm » Sat Sep 26, 2009 7:51 am
Strictly speaking the problem could be looked at as "the sum of all consecutive even integers from 2 to (n-1), inclusive".

first even number= 2
last even number = n-1

Average of all the even #s from first to last =
[(n-1)+2]*1/2 =(n+1)/2
(i.e (first + last)/2)

Total number of even integers from first to last =
1/2*[(n-1) - 2] + 1
= (n-1)/2
(i.e (last -first)/2 + 1)

Sum = (Average) * (Total number of even integers from first to last)

=>
(n+1)/2 * (n-1)/2 =79*80
=> n ~ SR(4*79*80)
n ~SR(4*80*80)
n~160

This takes a minute to do.

Well, we all know n=158.993711...

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by liciniudragos » Mon Jan 25, 2010 4:50 am
After some quick search in wikipedia This is what I found:

Sum of all even numbers up to n = N(N+1) where N is the number of consecutive even numbers

So: 79*80=N(N+1) therefore N=79 the last term of the sequence

the nth number value is expressed as n(N)=a+(N-1)*d where a = 1st term here 2 and d = increment here 2

n(79)=2+(79-1)*2 so n(79)=158 so the next odd number is 159

q.e.d:)

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by tailoc » Tue Sep 20, 2011 11:46 pm
I used formula: sum of consecutive even integers = [(last even integer + first even integer)/2]*[(last even integer - first even integer)/2 +1]
n is last odd integer in set =>n-1 is last even integer in set
2 is first even integer in set
So
(n-1+2)/2 * [(n-1-2)/2 +1] = 78*79
(n+1)/2 * [(n-3)/2 +1]= 78*79
(n+1)*(n-1)/2=78*79
(n+1)*(n-1)= 158*156
=>n=157

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by GMATGuruNY » Wed Sep 21, 2011 3:13 am
800guy wrote: The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
The solution is VERY quick if you realize that 79*80 represents the formula for the sum of evenly spaced integers:

Sum = number * average

Thus:
The number of even integers between 1 and n = 79.
The average of the biggest and the smallest = 80.

Since there are 79 even integers between 1 and n, the biggest = 2*79 = 158.
To confirm, the average of the biggest and the smallest = (158+2)/2 = 80.
Since the greatest even integer in the set is 158, and n is the next largest ODD integer, n = 159.
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by Newborg » Wed Mar 21, 2012 7:56 am
My solution:

Write the even integers between 1 and n (where n is odd) in two rows with opposite order:

ASCENDING ORDER: 2, 4, 6,..., (n-5), (n-3),(n-1)

DESCENDING ORDER: (n-1), (n-3), (n -5)..., 6, 4 , 2

Summing each term in the two rows gives:
(n+1) + (n+1) .... + (n+1) = (n-1)*(n-1)/2 (since there are (n-1)/2 terms)

This is twice the sum of the even numbers between 1 and n.
Therefore:
(n-1)*(n+1)/4 = 79 * 80
=>
(n-1)*(n+1) = (2*79) * (2*80) = 158 *160
=> n = 159

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by ronnie1985 » Wed Mar 21, 2012 9:03 am
2+4+6+...2n = 2*(1+2+3+...+n) = n*(n+1) = 79*80 => n = 79 or 80
If n = 79, number = 2*79+1 = 159
If n = 80, number = 2*80+1 = 161
Check: 159: 2+4+...+158 = 2*(1+2+...+79) = 79*80
Check: 161: 2+4+...+160 = 2*(1+2+...+80) = 80*81