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        riddles >> easy >> Distinct Division
        
(Message started by: Sir Col on Aug 29^th, 2007, 7:20am)

Title: Distinct Division
Post by Sir Col on Aug 29^th, 2007, 7:20am

Given that a, b, and c are different digits, how many distinct solutions exist in the quotient: baba / a = aca?

(Note that baba represents a 4-digit number and not the product a²b²)

Title: Re: Distinct Division
Post by Grimbal on Aug 29^th, 2007, 7:37am

[hide]
Let's write it: a·aca = baba

Considering the last digit, a can only be 1, 5, 6, and it is obviously not 1.

5*5c5 = 2525 + 50·c works only with c=0
6*6c6 = 3636 + 60·c works only with c=0

So, I only see (a,b,c) = (5,2,0) and (6,3,0)
[/hide]

Title: Re: Distinct Division
Post by towr on Aug 29^th, 2007, 8:25am

[hide]a * aca = baba
baba = 101 * ba

101 * ba = a * aca
for a < 101, 101 must divide aca, so then c = 0

so

ba = a * a

In decimal,that leaves us 3,6 and 2,5 for b,a
For other bases I'll have to take a better look. and of course for bases in which a can be greater than 101, the story can also be different[/hide]

Title: Re: Distinct Division
Post by towr on Aug 29^th, 2007, 8:39am

[hideb]All pairs a,b in (natural) number bases smaller or equal to 101

6: (3, 1), (4, 2)
10: (5, 2), (6, 3)
12: (4, 1), (9, 6)
14: (7, 3), (8, 4)
15: (6, 2), (10, 6)
18: (9, 4), (10, 5)
20: (5, 1), (16, 12)
21: (7, 2), (15, 10)
22: (11, 5), (12, 6)
24: (9, 3), (16, 10)
26: (13, 6), (14, 7)
28: (8, 2), (21, 15)
30: (6, 1), (10, 3), (15, 7), (16, 8), (21, 14), (25, 20)
33: (12, 4), (22, 14)
34: (17, 8), (18, 9)
35: (15, 6), (21, 12)
36: (9, 2), (28, 21)
38: (19, 9), (20, 10)
39: (13, 4), (27, 18)
40: (16, 6), (25, 15)
42: (7, 1), (15, 5), (21, 10), (22, 11), (28, 18), (36, 30)
44: (12, 3), (33, 24)
45: (10, 2), (36, 28)
46: (23, 11), (24, 12)
48: (16, 5), (33, 22)
50: (25, 12), (26, 13)
51: (18, 6), (34, 22)
52: (13, 3), (40, 30)
54: (27, 13), (28, 14)
55: (11, 2), (45, 36)
56: (8, 1), (49, 42)
57: (19, 6), (39, 26)
58: (29, 14), (30, 15)
60: (16, 4), (21, 7), (25, 10), (36, 21), (40, 26), (45, 33)
62: (31, 15), (32, 16)
63: (28, 12), (36, 20)
65: (26, 10), (40, 24)
66: (12, 2), (22, 7), (33, 16), (34, 17), (45, 30), (55, 45)
68: (17, 4), (52, 39)
69: (24, 8), (46, 30)
70: (15, 3), (21, 6), (35, 17), (36, 18), (50, 35), (56, 44)
72: (9, 1), (64, 56)
74: (37, 18), (38, 19)
75: (25, 8), (51, 34)
76: (20, 5), (57, 42)
77: (22, 6), (56, 40)
78: (13, 2), (27, 9), (39, 19), (40, 20), (52, 34), (66, 55)
80: (16, 3), (65, 52)
82: (41, 20), (42, 21)
84: (21, 5), (28, 9), (36, 15), (49, 28), (57, 38), (64, 48)
85: (35, 14), (51, 30)
86: (43, 21), (44, 22)
87: (30, 10), (58, 38)
88: (33, 12), (56, 35)
90: (10, 1), (36, 14), (45, 22), (46, 23), (55, 33), (81, 72)
91: (14, 2), (78, 66)
92: (24, 6), (69, 51)
93: (31, 10), (63, 42)
94: (47, 23), (48, 24)
95: (20, 4), (76, 60)
96: (33, 11), (64, 42)
98: (49, 24), (50, 25)
99: (45, 20), (55, 30)
100: (25, 6), (76, 57)
[/hideb]

Title: Re: Distinct Division
Post by brown_eye on Aug 31^st, 2007, 6:01am

reduce it to (10b)-(10/101)*c = a(a-1)...
that can be an approach ..... :P

Title: Re: Distinct Division
Post by Grimbal on Aug 31^st, 2007, 6:15am

It would be 10/101*ac, but it comes to the same.