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2+6+10+14+18+22+26+30+34+38=200

2(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)  = 200

Let’s simplify it and get ten odd numbers that sum to 100.

2(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)  = 200

(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)  = 200/2

(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)  = 100

Now, to rephrase the question, we have to pick five numbers from above set {1,3,5,7,9,11,13,15,17,19} and get a sum of 50. It’s same as question, I just halved it 🙂

Now, 50 is an even number. We have to pick any five odd numbers, add them, get an even sum i.e. 50.

Adding odd number of odd numbers will provide always an odd result (re-read) 🙂 Example {3,9,13,17,19} sum is 61. Remember (odd+odd = even; odd + even = odd).

Thus, picking any five numbers from the set {1,3,5,7,9,11,13,15,17,19} always give an odd sum and that’s contrary to asked sum. So there is no answer. Such a subset do not exist.

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