if _+_+_=30. then find out the values that will be fill the blank if the condition is that you can only take the odd numbers from 1 to 15 and same number can be used more than once.
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Answered by
2
The sum of
three odd numbers gives an odd number
So, the sum of 3 given odd numbers doesn’t results 30
So, it is an impossible factor
here we can't use 3! too
because 3! [3×2×1] results a even number 6
so we can’t use it.
If there will be an allowance of factorials, then the answers will be
3! + 15 + 9 =30
3! + 11 + 13 = 30
So, the sum of 3 given odd numbers doesn’t results 30
So, it is an impossible factor
here we can't use 3! too
because 3! [3×2×1] results a even number 6
so we can’t use it.
If there will be an allowance of factorials, then the answers will be
3! + 15 + 9 =30
3! + 11 + 13 = 30
Anonymous:
it is a wrong question
Answered by
1
we know that odd+odd=always even
and odd+odd+odd=always odd then
odd+odd+odd = 30 is not possible because 30 is a even number to make this possible we should have allowance of adding one even no. because odd+odd+even = even but we have to choose only odd no. 1 to 15 it means that it is an impossible factor
and odd+odd+odd=always odd then
odd+odd+odd = 30 is not possible because 30 is a even number to make this possible we should have allowance of adding one even no. because odd+odd+even = even but we have to choose only odd no. 1 to 15 it means that it is an impossible factor
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