Answer by Akshat Sinha:
Firstly I will show you it is not possible by direct sum, then I’ll demonstrate the different ways to do so.
ANSWER:- Not Possible by direct sum in base 10, because
You are viewing: Which 3 Odd Numbers Add Up To 30
If you take the numbers to be modulo 2, five odd numbers would sum up to 1, but 30 is 0 modulo 2
WHY:- Prerequisites:- Knowledge of basic Mathematics.
Number’s are odd numbers
1=2*0+1 3=2*1+1 5=2*2+1 . . . 13= 2*6+1 15= 2*7+1 Or we can say that 2*n+1 is an odd number
Now we can say that the number we have to prove
To prove:- x+y+z=30
For all x,y,z ∈ 2n+1 where n∈N(Natural numbers). Proof:-here’s no combination. It’s impossible. Every number in that list is odd.
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We know that any even number can be expressed as 2n where n is an integer. Any odd number can be expressed as 2n−1.
Let m, n, and k all be integers. Then the sum of three odd numbers can be written as: (2m−1)+(2n−1)+(2k−1) =2(m+n+k−1)−1.
Note that m+n+k−1 is another integer. If we call it N then the result is simply 2N−1 which, as we know is an odd number. Therefore the sum of three odd numbers is always an odd number.
Hence Proved.
In simple words, there do not have any odd numbers, where by adding any 3 of them we get 30 because
odd + odd + odd = (odd + odd) + odd = even + odd = odd (and 30 is even number)
Example:-
Let the number is 15,13 and 3 then 15=2*7+1 13=2*6+1 3=2*1+1 where 1,6 and 7 now 15+13+3=2*7+1+2*6+1+281+11 =2(7+6+1)+3 =2*14+3(24 is our ‘m’) and 2*14≠30. OR 15+13+3≠30.
CONCLUSION:- So by adding 3 odd numbers, we do not get 30. Is there any other way to do so? Yes… Read below..
2.) How Possible:-
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With using any function within an operator How Possible?:-
- Factorial method:-
3!+11+13=30 3!+9+15=30 3!+3!+3!+3!+3!=30 11+1+3!+3!+3!=30 5+7+3!+3!+3!=30 11+7+3!+3!+3!=30 3+5+7+9+3!=30 1+5+7+11+3!=30 1+1+3!+11+13=30 1+3+5+3!+15=30 and many more. Click here to Learn How to find Factorial of a program (Mirror Link)
- Changing Base method :-
[(3) + (3) + (3) + (3) + (3)] 5= [(6)+(6)+(6)+(6)+(6)]10= [30]10 (1)7+(5)7+(5)7+(5)7+(5)7=(30)710 + 10 + 10 = 30 in any odd base (seven, nine, eleven, and so on). Click here to learn_base_conversion (Mirror Link)
- Addition-Subtraction Method:- a+b+c=a+b+(x-y) where x-y=c
15+(13-11)+(9+7-5+3-1)=30 15+(13-11)+(9+7-3)=30 15+(13-9)+(7+3+1)=30 15+(13-9)+(7+5-1)=30 15+(13-7)+(5+3+1)=30 15+13+(3-1)=30 15+13+(5-3)=30 15+13+(7-5)=30 15+13+(9-7)=30 15+13+(9-5-3+1)=30 15+13+(11-9-7+5+3-1)=30 15+13+(11-9 )=30 15+13+(11-9+7-5-3+1)=30 15+13+(11-7-5+3)=30 15+13+(11-7-3+1)=30 15+13+(11-5-3-1)=30
- OTHER METHOD 1) 30 cm=(15 cm+13 cm)+ 1cm +(7 mm+3mm) 2)( 1+3)+(13+7) +6 (inverting 9 using mirror ) = 30 3) 11.3 + 15.7 + 3 = 30 (Using decimal) 4) 30 years = 15 years + 13 years + 1 year + 7 months + 5 months(USING Calendar Mathematics) 5) ((13+11)/3) + 1 + 5 + 7 + 9 = 30 (using numbers 1, 3, 5, 7, 9, 11, 13) 7) (9/3)+(15/5)+(13+1/7)+(11)+(11)=30 (Using Division Operator) =>3+3+2+11+11=30
- Commas were used instead of decimal points in the past history by Arabs and even in France. You will be surprised to hear that even today in some parts of the world. Commas are written instead of decimal unit or a dot!Using commas in place of decimal units we can easily come up with a solution equal to 30!For example: 13,7 + 15,3 + 1 = 30 !
Can you solve _ + _ + _ = 30 using 1, 3, 5, 7, 9, 11, 13 and 15?
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