Question

addition of 20 in six boxes not by 0 and digit 0to9 and one digit not can be repeat
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Answer 1

Answer:

First, I’m going to assume that “number” means a non negative integer. There are infinite combinations if you allow negatives, and if you allow rational numbers, and I don’t even want to think about irrationals and complex numbers.

Since the numbers are all different, the first five numbers must be at a minimum

0+1+2+3+4=10. But then the next number would have to be 10. That’s a number but not a digit, so if we’re talking only digits, this solution is out.

The series of numbers

1+2+3+4+5=15, meaning the next digit would have to be 5, and substituting higher numbers for the 5 mean needing lower numbers for the final digit, those lower numbers will similarly be already taken. So no series without a zero can work. If zero is ruled out, there are no solutions in the positive integers.

With a zero, though, there are more combinations that might work

with a start of 0+1+2+3, the combinations 5,9 and 6,8 also work. 7,7 would require a repeat, and 8,5 9,5 and 10,4 are the same as solutions we already have

0+2+3+4+5=14, +6=20

No other combinations with 0+2+3+4 work, that is the only solution without a 1.

0+1+3+4+5+6=14, meaning the next digit must be 6, so 2 must be included in any series0+1+2+4+5+6=18, meaning the next digit must be 2, so 3 must be included in any series

we’ve already addressed all possible combinations which start 0,1,2,3

so to sum up:

(0,1,2,3,4,10)

(0,1,2,3,5,9)

(0,1,2,3,6,8)

(0,2,3,4,5,6)

If 0 is not an allowed number, there are no solutions in the positive integers.

hope it helps u

Answer 2

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