Problem Solving

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:)

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

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.

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

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
Hence, 159 is the answer

159 is the correct answer. see the attached solution for the algebraic method.

ttwang56 wrote: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

question says....SUM of all even numbers from 1 to n-1 is 79*80

so.. if n=159, n-1 = 158

We need to add..... 2+4+6+.....158
By A.P. formula

Sum = n/2(first term + last term), Here n = number of terms, not the n from our question.(which is 79)

Sum =79*(2+158) / 2
=79*160/2
=79*80

I hope you got it.....

My ans
n is 159