Sorry for this, and nothing against programming, but for me it was a question of honour.
Let's call the numbers ABC and DEF.
We are searching for the values of A, B, C, D, E and F, in the way that (ABC x DEF) = (AD x BE x CF).
(ABC x DEF) is congruent to (CxF) (mod 10).
(AD x BE x CF) is congruent to (DxExF) (mod 10).
So, (CxF) is congruent to (DxExF) (mod 10).
Dividing by F, C is congruent to DxE (mod 10/F).
Since 10/F must be an integer, F is equal to 5 or to 2.
a) F = 5
C is congruent to DxE (mod 2), and can be 2, 3, 4, 6 or 7.
C can't be odd, because the product DxE would have at least one even number, so C would be congruent to 1 (mod 2) and DxE would be congruent to 0 (mod 2).
Therefore, C must be even.
For C even (2, 4 or 6), and so congruent to 0 mod 2, D and E cannot be both odd (their product would be congruent to 1 mod 2). So, at least one of them (D,E) must be even.
a1) For C = 2, the product (AD x BE x 25) is a multiple of 25, and so must be the product (AB2 x DE5). AB2 is not multiple of 5, so DE5 must be multiple of 25. But a multiple of 25 ends in 00, 25, 50 or 75. So (C is already 2), E must be 7, and since D could not be odd too, D = 4 or 6, wich yields DE5 = 475 or 675.
For DE5 = 475, AB2 = 362 or 632. Evaluating a few multiplications, we see that DE5 = 475 can be eliminated.
For DE5 = 675, AB2 = 342 or 432. And, also, with a few multiplications, we see that DE5 = 675 can be eliminated.
Therefore, C is not 2.
a2) For C = 4, the product (AD x BE x 45) is a multiple of 45, and so must be the product (AB4 x DE5).
45 = 3 x 3 x 5.
Since AB4 is not a multiple of 5, we have : a21) DE5 is a multiple of 45; or a22) AB4 is a multiple of 3 or 9 (since DE5 is already a multiple of 5).
a21) DE5 is a multiple of 45.
The unique multiple of 45 that serves is 765 = 17 x 45.
Then, AB4 would be 234 or 324. Evaluating the necessary products, we eliminate this case.
a22) AB4 is a multiple of 3 or 9.
So, (A+B) must sum to 5, and (A,B) = (2,3) or (3,2), and (D,E) = (6,7) or (7,6). AB4 would be 234 or 324 and DE5 would be 675 or 765. This case is almost equal to case a21. We need only to test 234 and 324 against 675, and we see that we can eliminate this case.
Therefore, C is not 4.
a3) For C = 6, the product (AD x BE x 65) is a multiple of 65, so must be the product (AB6 x DE5).
65 = 5 x 13
Since AB6 is not a multiple of 5, it must be a multiple of 13, and ending in 6, the possible factor must end in 2. Testing 12, 22, 32...72, we eliminate this case.
And we conclude that F is not equal to 5.
Therefore, F = 2.
C is congruent to DxE (mod 5).
C could be 3, 4, 5, 6 or 7.
b1) C = 3. (D,E) could be (4,5,6,7). (DxE) must be 8 or 13 or 18 or 23 or 28 or 33 or 38. We found an unique pair valid for (D,E), that is (4,7) [or (7,4)].
So, (A,B) = (5,6) or (6,5), and we have 563x472 or 563x742 or 653x472 or 653x742. Evaluating these cases, we eliminate all of them.
b2) C = 4. (D,E) could be (3,5,6,7). (DxE) must be 9 or 14 or 19 or 24 or 29 or 34 or 39, and we found no values for (D,E).
b3) C = 5. (D,E) could be (3,4,6,7). (DxE) must be congruent to 0 (mod 5), and we found no values for (D,E).
b4) C = 6. (D,E) could be (3,4,5,7). (DxE) must be congruent to 1 (mod 5), and the only cases is (D,E)=(3,7) [or (7,3)].
So, (A,B) = (4,5) or (5,4), and we have 456x372 or 456x732 or 546x372 or 546x732. And we eliminate all of them with a few multiplications.
b5) C = 7. (D,E) could be (3,4,5,6). (DxE) must be congruent to 2 (mod 5), and the only pair that serves is (3,4) [or (4,3)].
So, (A,B) = (5,6) or (6,5), and we have 567x342 or 567x432 or 657x342 or 657x432.
With a few multiplications, we arrive at the unique solution, that is 567x432 = 54x63x72=244,944.
Perhaps surprisingly, or not, all the numbers are multiples of 9.
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Posted by pcbouhid
on 2005-03-30 19:48:00 |
full solution - no programming
