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2006-09-18 17:15:54 · 5 answers · asked by Jill W 1 in Science & Mathematics Mathematics

Sorry, I didn't explain this well. The numbers are in three columns and three rows, added as three three-digit numbers.

2006-09-18 17:23:53 · update #1

5 answers

I´ve tried and the shortest to 777 was 774, like this:

147
258
369
---------
774

The 1st row has to be 1 2 3 (the smallest); if you exchange algarisms of the 2d. row with the 1st., in order to have the partial result of 7 in the first; well ...it is late (in my country!). I give up.
I answer that, in the way it is formulated, it is impossible, an that the closest to 777 is 774. O. K. ? Let´s see if somebody got it.
Wow! I see now that the comment before mine says it has to be a multiple of 9! And 774 is! Good! it was fun!

2006-09-18 18:42:41 · answer #1 · answered by Vovó (Grandma) 7 · 0 0

I will prove it is impossible. Any sum of three three-digit numbers can be divided by 9. Your number, 777, is divisible by 3 but not by 9.

Prove of my statement: suppose the three numbers you add are [abc], [def] and [ghi], where a through i are a permutation of 1 through 9. Now [abc] stands for a hundreds, b tens and c units, that is,

[abc] = 100 a + 10 b + c

and so on. Adding the three numbers, we get

[abc] + [def] + [ghi] =
... = 100 (a + d + g) + 10 (b + e + h) + (c + f + i)
... = 99 (a + d + g) + (a + d + g) + 9 (b + e + h) + (b + e + h) + (c + f + i)
... = 9 * X + (a + b + c + d + e + f + g + h + i)
... = 9 * X + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
... = 9 * X + 45 = 9 * (X + 5)

where X = 11 (a + d + g) + (b + e + h).

This proves that the sum is a multiple of 9. The proof works for any addition or subtraction of arbitrary-length numbers in which each of the digits is used precisely once.

2006-09-19 00:47:12 · answer #2 · answered by dutch_prof 4 · 0 0

98 + 7 + 654 – 3 + 21 = 777

All numbers from one to nine were used once. Yes, sum includes subtracting numbers because you can "add" negative numbers.

[edit] With your new restriction, that you need to make three 3-digit numbers, then you pretty much need to have numbers starting with 1, 2, 3 (1##, 2##, and 3##) to allow the tens and ones to add to 177 (600 + 177 = 777). That simplifies the problem to finding a sum of 177 from three 2-digit numbers using the numbers 4, 5, 6, 7, 8, 9. The closest I could get are 174, from 47+58+69 (so 147 + 258 + 369 = 774), and 183, from 46+58+79 (146 + 258 + 379 = 783). Sorry, no can do.

2006-09-19 00:24:48 · answer #3 · answered by drunksage 2 · 0 0

Certainly not - if you add them up, you won't get anywhere nearr 777!

2006-09-19 00:18:55 · answer #4 · answered by yo 2 · 0 0

i'am sure you wont get it in a logical manner

2006-09-19 00:24:47 · answer #5 · answered by Sai♥Pranav 3 · 0 0

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