Problem Solving

banona wrote:I am sorry alpha_gmat and 800guy

you gave the same answer as in the word document of " tough math problems" and it's wrong. n is not 79. In fact, we should check the accuracy of the answers sometimes.

I got the same answer following the same path as you (amitamit2020)
In fact, I think n= 159 ( and the trick is that people try to have something like (n+1)*n at the end, so that it would be possible to guess one n is;
now, to verify, if you plug the value of n in the the sumof even numbers you will find that :
if n = 79
2+4+6+........+78 = 2*( 1+2+3+4+........38) = 2*(39/2)*38 =38 * 39

Now, if n= 159
2+4+6+........+158 = 2( 1+2+3+4+........79) = 2*(80/2)*79 = 79*80

yup, you're right--i do provide the "OA" that's listed in the diff math questions doc. i post these questions to promote discussion on these forums...

good catch on this one

That' great 800guy,
You've reached your goal making us thinking and discussing around some problems,
we are waiting for your next question

Thanks

[quote="aim-wsc''] by now one can get the TREND the pattern...
add 79 & 80.
thats it.[/quote]
so mine was correct then :mrgreen:

btw 800guy is doing great job i must tell you about this 8)
his math question series helped many of our members.

thanks 800guy

thanks

shouldn't n = 157?

if n = 159 then the way to check would be 158+2 = 160. 160 would be able to added up all the with 4+156, etc. 160 doesn't go into the 6320 (78*80) evenly. However, if n = 157, then 2+156= 158. 158 goes evenly into 6320 with a solution of 40.

why can't u solve it by factoring 80 to being 2 x 40 and multiplying the 2 to the 79?

don't know if i explained this clearly or not.

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....

80 + 79 = 159

Can this problem be solved this way?

Answer n = 159.

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

Pardon the typo above. It should be 4X3

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

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.........

nice question 800guy

What difficulty level is this sort of question? approximation of 650?

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.

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...

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

link: https://wiki.answers.com/Q/What_is_the_s ... en_numbers

q.e.d:)