(1)Take a 2-digit number and multiply the two digits together. Repeat with each product until a single digit is the result. For most starting numbers, no more than three of these steps are needed to arrive at the final single digit. (For example: 57: 5*7=35: 3*5=15: 1*5=5). Find one that takes four steps.
(2)Start with a 3-digit number. Choose any one of the digits and remove it to make a 2-digit number. Multiply the 2-digit number by the digit you removed. Eventually, you will get a 2-digit number, which will yield a single digit, as above. What number should you start with, and what steps should you take to find the longest series of steps you can take to arrive at that final single digit
DECLARE SUB cntnu ()
CLEAR , , 25000
CLS
FOR n = 10 TO 99
p = n: PRINT p;
WHILE p > 9
m1 = p 10: m2 = p MOD 10
p = m1 * m2
PRINT p;
WEND
PRINT
IF n = 54 THEN DO: LOOP UNTIL INKEY$ > "": PRINT
NEXT
DO: LOOP UNTIL INKEY$ > "": PRINT
DIM SHARED h(50), ct, max, hmax(50), hw(50) AS STRING
FOR n = 100 TO 999
h(1) = n
ct = 1
cntnu
NEXT
SUB cntnu
p = h(ct)
d1 = p 100: d2 = (p 10) MOD 10: d3 = p MOD 10
p = d1 * (10 * d2 + d3): ct = ct + 1: h(ct) = p: hw(ct) = "h"
IF p < 10 AND d1 > 0 THEN
GOSUB checkmax
ELSE
IF p > 0 THEN cntnu
END IF
ct = ct - 1
p = d2 * (10 * d1 + d3): ct = ct + 1: h(ct) = p: hw(ct) = "t"
IF p < 10 AND (d1 > 0 OR d2 > 0) THEN
GOSUB checkmax
ELSE
IF p > 0 THEN cntnu
END IF
ct = ct - 1
p = d3 * (10 * d1 + d2): ct = ct + 1: h(ct) = p: hw(ct) = "u"
IF p < 10 AND (d1 > 0 OR d2 > 0) THEN
GOSUB checkmax
ELSE
IF p > 0 THEN cntnu
END IF
ct = ct - 1
EXIT SUB
checkmax:
IF ct >= max THEN
match = 1
FOR i = 1 TO ct
IF h(i) <> hmax(i) THEN match = 0: EXIT FOR
NEXT
IF ct > max THEN match = 0
IF match = 0 THEN
FOR i = 1 TO ct
IF i > 1 THEN PRINT hw(i);
PRINT h(i);
hmax(i) = h(i)
NEXT
PRINT
END IF
max = ct
END IF
RETURN
END SUB
finds for part 1:
10 0
11 1
12 2
13 3
14 4
15 5
16 6
17 7
18 8
19 9
20 0
21 2
22 4
23 6
24 8
25 10 0
26 12 2
27 14 4
28 16 6
29 18 8
30 0
31 3
32 6
33 9
34 12 2
35 15 5
36 18 8
37 21 2
38 24 8
39 27 14 4
40 0
41 4
42 8
43 12 2
44 16 6
45 20 0
46 24 8
47 28 16 6
48 32 6
49 36 18 8
50 0
51 5
52 10 0
53 15 5
54 20 0
55 25 10 0
56 30 0
57 35 15 5
58 40 0
59 45 20 0
60 0
61 6
62 12 2
63 18 8
64 24 8
65 30 0
66 36 18 8
67 42 8
68 48 32 6
69 54 20 0
70 0
71 7
72 14 4
73 21 2
74 28 16 6
75 35 15 5
76 42 8
77 49 36 18 8
78 56 30 0
79 63 18 8
80 0
81 8
82 16 6
83 24 8
84 32 6
85 40 0
86 48 32 6
87 56 30 0
88 64 24 8
89 72 14 4
90 0
91 9
92 18 8
93 27 14 4
94 36 18 8
95 45 20 0
96 54 20 0
97 63 18 8
98 72 14 4
99 81 8
so 77, 49, 36, 18, 8 is the sequence sought by part 1.
The longest sequence found for part 2 has 10 steps:
878 u 696 t 594 t 486 t 368 u 288 t 224 u 88 t 64 t 24 t 8
887 h 696 t 594 t 486 t 368 u 288 t 224 u 88 t 64 t 24 t 8
898 h 784 t 592 t 468 u 368 u 288 t 224 u 88 t 64 t 24 t 8
988 t 784 t 592 t 468 u 368 u 288 t 224 u 88 t 64 t 24 t 8
999 h 891 h 728 u 576 t 392 h 276 t 182 t 96 t 54 t 20 t 0
999 h 891 h 728 u 576 t 392 t 288 t 224 u 88 t 64 t 24 t 8
999 t 891 h 728 u 576 t 392 h 276 t 182 t 96 t 54 t 20 t 0
999 t 891 h 728 u 576 t 392 t 288 t 224 u 88 t 64 t 24 t 8
999 u 891 h 728 u 576 t 392 h 276 t 182 t 96 t 54 t 20 t 0
999 u 891 h 728 u 576 t 392 t 288 t 224 u 88 t 64 t 24 t 8
where the h, t or u specifies whether the hundreds, tens or units position was the one pulled out as the separate multiplier. A sequence is shown only if it has one or more numbers different from another of the sequences, so that, for example, when starting with 887 it did not matter whether you started with removal of the hundreds or the tens position. With 2-digit number of course it never matters.
Added note:
I see the two sequences for 999 were multiplied by 3 as the check for duplicates was only on consecutive solutions of the same maximum length, and since there were two real solutions, they separated the sequences that were duplicates.
Edited on November 29, 2013, 5:23 pm
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Posted by Charlie
on 2013-11-29 17:18:15 |
computer solution
