Forum |

An even number

posted by Ragib
- March 07 2011 09:58:46 PM

Find the lowest even number which can be explained as sum of 11 consecutive numbers,5 consecutive numbers,3 consecutive numbers.

Reply by Caprico
- March 10 2011 10:53:40 AM

330?

It can be written as the sum of 25-35

It can be written as the sum of 64-68

It can be written as the sum of 109-111

If you want to know how I calculated it, I'll be happy to elaborate in the next comment. Just in case you want to know.

It can be written as the sum of 25-35

It can be written as the sum of 64-68

It can be written as the sum of 109-111

If you want to know how I calculated it, I'll be happy to elaborate in the next comment. Just in case you want to know.

Reply by Drakelar
- March 11 2011 02:36:00 AM

Elaborate please

Reply by Caprico
- March 11 2011 08:04:30 AM

Keep writing this, or else you might get confused.

Let the 11 consecutive numbers be (x-5), (x-4).... (x).... (x+4), (x+5). On addition, they'll give you 11x. (Yes, I chose the numbers deliberately in such a manner, that they would conveniently cancel out.

Do the same for 5, and 3 consecutive numbers with y and z. You'll get 5y, and 3z.

Now all these represent the same number, i.e.

11x = 5y = 3z

Hence the number must be a common multiple of all three, 11, 5 and 3. That is 165. But the required number is even. So just multiply 165 with 2, and you get 330.

If you want, just equate this value with the above three equations, and you'll have the values of x, y and z.

Then the summations are from

1) (x-5) to (x+5)

2) (y-2) to (y+2)

3) (z-1) to (z+1)

And hence the problem is solved! :D

Let the 11 consecutive numbers be (x-5), (x-4).... (x).... (x+4), (x+5). On addition, they'll give you 11x. (Yes, I chose the numbers deliberately in such a manner, that they would conveniently cancel out.

Do the same for 5, and 3 consecutive numbers with y and z. You'll get 5y, and 3z.

Now all these represent the same number, i.e.

11x = 5y = 3z

Hence the number must be a common multiple of all three, 11, 5 and 3. That is 165. But the required number is even. So just multiply 165 with 2, and you get 330.

If you want, just equate this value with the above three equations, and you'll have the values of x, y and z.

Then the summations are from

1) (x-5) to (x+5)

2) (y-2) to (y+2)

3) (z-1) to (z+1)

And hence the problem is solved! :D

Reply by Ragib
- March 11 2011 09:52:54 AM

good logic,caprico

Reply by Drakelar
- March 13 2011 11:53:10 PM

I agree

Reply by Caprico
- March 14 2011 01:25:59 AM

Thank you!

To post a response, simply log in with your Google Account.

Log in | Desktop Site |