Problem Solving

I failed this one while preparing. Solved it an hour later after revising. It was along shot though and wasted too much time. Woul dlike if some of yuo can solve it in a timely manner.

IF n is an odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, what is the value of n ?

1)79

2)81

3)83

4)157

5)159

The problem should read as follows:

sapuna wrote: If n is an odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, what is the value of n ?

1)79

2)81

3)83

4)157

5)159

For any set of evenly spaced integers:
Number of integers = (biggest - smallest)/interval + 1, where the interval is the distance between one integer and the next.
Average = (biggest + smallest)/2.
Sum = (number of integers)(average).

We can PLUG IN THE ANSWERS, which represent the value of n.
When the correct answer choice is plugged in, the sum of the even integers between 1 and n = 79*80.

Answer choice D: 157
Since n=157, we must calculate the sum of the even integers between 1 and 157:
{2, 4, 6....152, 154, 156}.

In the set above:
Smallest = 2.
Biggest = 156.
Since all of the integers are even, the INTERVAL -- the distance between one integer and the next -- is 2.

Thus:
Number of integers = (156-2)/2 + 1 = 78.
Average = (156+2)/2 = 79.
Sum = 78*79.
Too small.
For the sum increase to 79*80, the value of n must increase.

The correct answer is E.

Answer choice E: 159
Since n=159, we must calculate the sum of the even integers between 1 and 159:
{2, 4, 6....154, 156, 158}.

In the set above:
Smallest = 2.
Biggest = 158.
Since all of the integers are even, the INTERVAL -- the distance between one integer and the next -- is 2.

Thus:
Number of integers = (158-2)/2 + 1 = 79.
Average = (158+2)/2 = 80.
Sum = 79*80.
Success!

sapuna wrote:I failed this one while preparing. Solved it an hour later after revising. It was along shot though and wasted too much time. Woul dlike if some of yuo can solve it in a timely manner.

IF n is an odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 90, what is the value of n ?

1)79

2)81

3)83

4)157

5)159

Sum of all even numbers from 1 till n = 2+4+6+8+.....+(n-1)
Product of 79 x 80

CONCEPT :
Sum of First a consecutive integers (from 1 till a) is given by = (1/2)*a*(a+1)
Sum of First a Consecutive even Integers (from 1 till 2a) is given by = a*(a+1)

here 79 x 80 = a (a+1)
i.e. a = 79 i.e. 79 Consecutive Even Integers
Which will fall in first 2x79 = 158 integers
but since n is odd therefor n=158+1 = 159

i.e. n = 159

Answer: Option E

Aye, I did it in a simular way. If somebody still wants it , I can give my solution as well. It`s more or less the same. Just without backdooring.

p.s. Sorry for the typos.

n = odd

No. of EVEn Numbers = (n-1)/2

Sum of (n-1)/2 Even Numbers = (n-1)/2 * ( (n-1)/2 + 1)

(n-1)/2 * ( (n-1)/2 + 1) = 79 * 80

(n-1)/2 * (n+1)/2 = 79 * 80
(n-1)(n+1) = 158 * 160

if we compare RHS & LHS then,

n = 159

I did it in a super simular way , just slightly different in the end , or I`d like to think so

As you guys pointed out , number of integers = (biggest - smallest )interval + 1

since were talking about even integers , the biggest even integer should be n -1 and the smallest 2

therefore number = (n-1 - 2)/ 2 + 1 = (n-3)/2 + 1 = (n - 1)/2

the average is (biggest + smallest ) :2 => n-1 + 2 all devided by 2 = (n + 1)/2

Sum = average x number

79.80 = (n-1)(n+1)/2

79.160 = (n-1)(n+1)

n+1 will always > n - 1 . Therefore N+1= 160 = > n = 159