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 Thirds and Differences (Posted on 2005-07-21)

Arrange the digits 1-9 so that the first 3 form a number that is 1/3 of the number formed by the last three and the three digits in the middle form a number that is the difference between them.

The numbers are as they appear, not summed.

 No Solution Yet Submitted by Erik O. Rating: 3.0000 (3 votes)

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 solution - no computers ! | Comment 3 of 9 |

N = abc,def,ghi

3* abc = ghi,  and def = (ghi - abc)-----> ghi = (def + abc)

3 * abc = (def + abc) -----> def = 2*abc------> a < d.

3*abc = ghi ----> a <=3 -----> d <= 7. And g >= 3.

a + d = g ------->(a,d) = (1,2), (1,3),(1,4),(1,5),(1,6),(1,7),(2,3),(2,4),(2,5),(2,6),(2,7),(3,4),(3,5),(3,6),(3,7).

N = (1bc,2ef,3hi), (1bc,2ef,4hi), (1bc,3ef,4hi), (1bc,3ef,5hi),  (1bc,4ef,6hi), (2bc,3ef,4hi), (2bc,3ef,5hi), (2bc,4ef,6hi), (2bc,4ef,7hi), (2bc,5ef,7hi), (2bc,5ef,8hi), (2bc,6ef,9hi), (3bc,4ef,5hi), (3bc,4ef,6hi), (3bc,5ef,6hi), (3bc,5ef,7hi), (3bc,6ef,8hi), (3bc,6ef,9hi).

Since 3*abc = ghi, the possibilities are reduced to :

N = (1bc,2ef,3hi), (1bc,2ef,4hi), (1bc,3ef,4hi), (1bc,3ef,5hi), (2bc,4ef,6hi), (2bc,4ef,7hi), (2bc,5ef,8hi).

I) N = 1bc,2ef,3hi.

3*1bc = 3hi, so b =< 3, but we already have 1, 2 and 3.

II) N = 1bc,2ef,4hi.

3*1bc = 4hi, so b = 3 or 4 or 5 or 6.

b = 3 -----> 3*13c = 4hi ----->to (3*3+x) carry 1, h would be 0 or 1 ---->impossible.

b = 4 ----> 3*14c = 4hi -----> h would b 2, 3 or 4 ---->imp.

b = 5 ----> 3*15c = 4hi -----> h = 6 or 7 (possibles)

3*15c = 46i -----> (c,i) => no values.

3*15c = 47i -----> (c,i) = (8,4) --->impossible.

b = 6 ----> 3*16c = 4hi -----> h = 8 or 9

3*16c = 48i ------> (c,i) = (9,7)

N = 169,2ef,487

This leaves 3 and 5 for e and f.

169 + 487 = 656

656 / 2 = 328.  impossible.

III)  N = 1bc,3ef,4hi.

3*1bc = 4hi, so b = 5 or 6.

b = 5 ----> 3*15c = 4hi ----->h = 6 or 7

3*15c = 46i -----> (c,i) no values (3*c could only    carry 1.

3*15c = 47i -----> (c,i) no values.

b = 6 ----> 3*16c = 4hi -----> h = 8 or 9.

3*16c = 48i ------> (c,i) = (9,7)

N = 169,3ef,487.

169 + 487 = 656

656 / 2 = 328 impossible.

3*16c = 49i ------->(c,i) no values.

IV) N = 1bc,3ef,5hi

3*1bc = 5hi, so b = 7 or 8 or 9.

b = 7 ----> 3*17c = 5hi -----> h = 2.

3*17c = 52i ------> (c,i) = (6,8)

3*176 = 528------->N = 176,3ef,528

176 + 528 = 704

704 / 2 = 352 impossible.

b = 8 ----> 3*18c = 5hi -----> h = 6

3*18c = 56i ------> (c,i) = (7,9)

3*187 = 569 -------> N = 187,3ef,569

187 + 569 = 756

756 / 2 = 358 impossible.

b = 9 ----> 3*19c = 5hi ------> h = 7 or 8

h = 7...... 3*19c = 57i....(c,i) = (2,6)

N = 192,3ef,576

192 + 576 = 768     768/2 = 384

N = 192,384,576

And I`ll stop here, since I found a solution.

 Posted by pcbouhid on 2005-07-21 16:52:52

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