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How do you get to 100 by adding 1-9 only plus and usind each number only once?
Wiki User
∙ 11y ago
You cannot and here is an outline of the proof. It depends on two things:
- digital root (DR): The digital root of an integer is the sum of all its digits. If the result is greater than 9 you calculate the DR of the sum and continue until you get below 10.
- divisibility rule for 9: An integer is divisible by 9 if and only if its DR is 9 (or 0).
The digital root of any two-digit number is the sum of their digital roots. That is DR(42) = 4 + 2 = DR(4) + DR(2).
That result can be generalised and used to show that, however you try to group the 9 digits before summing (for example 12 + 38 + 4 + 5 + 967), the DR is always the same as 1+2+3+4+5+6+7+8+9 = 45 = 9
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In the above example,
DR(12) + DR(38) + DR(4) + DR(5) + DR(967)
= DR(1)+DR(2)+DR(3)+DR(8)+DR(4)+DR(5)+DR(9)+DR(6)+DR(7)
= DR(1)+DR(2)+DR(3)+DR(4)+DR(5)+DR(6)+DR(7)+DR(8)+DR(9) - on reordering
= 1+2+3+4+5+6+7+8+9
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So if you use the digits only once, DR(Sum) = 9 and so, by the divisibility rule, the number must be divisible by 9.
But 100 is not divisible by 9. So the sum cannot be 100.
Wiki User
∙ 11y ago
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