Question

1+3+5+7+9+11+1+15+17+19=100 then out of this which 5 numbers make 50 .some logic working?

Answer 1

thats 4 numbers

Answer 2

all the numbers are odd ... 5 odd numbers = odd and 50 is even

Answer 3

right!!

Answer 4

/me agrees with @experimentX & waves hello to the new person ^_^

Answer 5

As the numbers are all odd numbers then 50 cannot be formed by adding 5 different odd numbers

Answer 6

If other mathematical operators are allowed (such as division), then
19 + 17 + 9 + (15 ÷ 3) = 45 + 5 = 50

Answer 7

I like this one. @experimentX got straight to the point!

You can construct the nth even number using this formula: \[2*n\]where n is an integer

You can then construct the nth odd number using this formula: \[2*n + 1\]again where n is an integer

In the problem, you are given 10 odd numbers,
\[(2*n_1+1,2*n_2+1,2*n_3+1, ..., 2*n_{10}+1)\]These numbers are summed to yield 100.

With some manipulation you have \[(2*n_1 + 2*n_2 + ... + 2*n_{10} + 1 + 1+ 1 + 1 + 1 + ... + 1)\]which is just separating the 1's from the \[2*n_i's\] so you have 10 2*ni factors and 10 1's/

Now, \[2*n_1 + ... + 2*n_{10} + 10 = 100\]is the original equation abstracted.

Your next goal is take any 5 of the 10 odd numbers and get the sum of 50.
\[((2*n_1+1) + (2*n_2+1) + (2*n_3+1) + (2*n_4 +1) + (2*n_5+1)) = 50\]
\[((2*n_1)+ (2*n_2) + (2*n_3) + (2*n_4) + (2*n_5) + 5) = 50\]
\[((2*n_1)+ (2*n_2) + (2*n_3) + (2*n_4) + (2*n_5) + 5 - 5 = 50 -5\]
\[((2*n_1)+ (2*n_2) + (2*n_3) + (2*n_4) + (2*n_5) = 45\]
Now this looks impossible, you are trying to sum 5 even numbers now that will equal an odd number, and an even number added to another even number yields another even number \[2*n_1 + 2*n_2 = even = 2*(n_1 + n_2)\]where \[n_1, n_2\]are integers