- + - + - + - + - =30 use online positive odd number you can repeat them but only 5 numbers
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Can you add 5 odd numbers to get 30?
ANSWER WIKI
No, unless the context is changed (e.g. in modular arithmetic or in a different base) because addition of 5 odd numbers cannot result in an even number.
100+ ANSWERS

Anupam Kumar, Aspiring actuary , reads poetry
Updated Mar 18, 2018
Originally Answered: _+_+_=30,using 1,3,5,7,9,11,13,15.you can repeat the numbers?
Sum of three or five odd number can't be an even number , is a simple high school mathematics . However on specific assumption we can reach to a solution . We all work out numerics in decimal numeral system , here I am taking an assumption of working out this problem in Numeral system with base 9 (radix=9) . So in our base 9 numeral system
1=1
3=3
5=5
7=7
9=9 Radix - Wikipedia Base 10 has 10 unique digits: [0,9]. Base 9 has 9 unique digits: [0,8]
11=1*9^1 +1*9^0 =10
13=1*9^1 + 3*9^0 =12
15 = 1*9^1 + 5*9^0 = 14
30 = 3*9^1 + 0*9^0 = 27
So our problem becomes :
__ + __ + __ = 27
where __ can be filled by 1,3, 5,7,9,10,12,14,27
which leads us to the solution
1 + 12 + 14 =27 .
Now we transform our solution back to the base 10 notation :
1+13+15 =30 , where 1 ,13,15,30 are taking its numeric value according to base 9 numeral system .
Similar logic can be used not five numbers.
ANSWER WIKI
No, unless the context is changed (e.g. in modular arithmetic or in a different base) because addition of 5 odd numbers cannot result in an even number.
100+ ANSWERS

Anupam Kumar, Aspiring actuary , reads poetry
Updated Mar 18, 2018
Originally Answered: _+_+_=30,using 1,3,5,7,9,11,13,15.you can repeat the numbers?
Sum of three or five odd number can't be an even number , is a simple high school mathematics . However on specific assumption we can reach to a solution . We all work out numerics in decimal numeral system , here I am taking an assumption of working out this problem in Numeral system with base 9 (radix=9) . So in our base 9 numeral system
1=1
3=3
5=5
7=7
9=9 Radix - Wikipedia Base 10 has 10 unique digits: [0,9]. Base 9 has 9 unique digits: [0,8]
11=1*9^1 +1*9^0 =10
13=1*9^1 + 3*9^0 =12
15 = 1*9^1 + 5*9^0 = 14
30 = 3*9^1 + 0*9^0 = 27
So our problem becomes :
__ + __ + __ = 27
where __ can be filled by 1,3, 5,7,9,10,12,14,27
which leads us to the solution
1 + 12 + 14 =27 .
Now we transform our solution back to the base 10 notation :
1+13+15 =30 , where 1 ,13,15,30 are taking its numeric value according to base 9 numeral system .
Similar logic can be used not five numbers.
Answered by
0
10+10+5+4+1=30
hope its help
hope its help
meghasg2004:
10 is even
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