The question is:
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$?
My answer to this question is $63$ but the right answer is $64$. I don't know why it is $64$? I need some assistance.
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$
combinatoricselementary-number-theory
Related Solutions
There are useful general techniques that apply to problems like these.
With distinct numbers, to do this fast, we should flip the picture around with the idea of Young diagrams. Three rows of different lengths, totalling 15 boxes is the same as a bunch of columns of lengths 1,2,3 with at least one of each length. Here's an example diagram for 10 (as opposed to 15) from Wikipedia:
Then we can break things down not by smallest number, but by number of threes.
- Five 3s? No, you need at least one 1 and one 2.
- Four 3s? 15-4*3=3, which you can only get as 1+2 with 1s and 2s. Thus, 1 possibility.
- Three 3s? Then we need to make 6 out of 2s and 1s, which can be done with two 2s or one 2: 2 possibilities.
- Two 3s? We need to make 9, so anywhere from one 2 to four 2s would be fine. So 4 possibilities.
- One 3? Then we need to make 12 out of 2s and 1s. We could take anywhere from one 2 to five 2s, so 5 possibilities.
- No 3s? Not allowed; we need at least one of each.
1+2+4+5=12, which is the answer if "whole numbers" means positive integers, as I think it does in this context.
And if "nonnegative integers" was meant, then we also need to include two-row Young diagrams. How can you build 15 out of 2s and 1s alone? You could have anywhere from one 2 to seven 2s, so you get 7 new things, so 12+7=19, as John found.
As an aside, if you weren't restricted to 2-3 minutes, and knew about generating functions, you could take this idea quite far.
If we didn't have the condition that the numbers were distinct, then the problem reduces to stars and bars: If the numbers $a$ $b$ and $c$ have to be positive integers, then $a-1$, $b-1$, and $c-1$ are nonnegative integers that sum to $15-3=12$, so that two bars and twelve stars should do it: ${14}\choose{2}$.
I know you guys love your maths (and I do too) but, for something as finite as this (i.e., not something that would consume years of compute time), you can test it with an exhaustive computer program, one that takes about three thousandths of a second on my desktop:
#include <stdio.h>
int main (int argc, char *argv[]) {
int quantN[200] = {0};
for (int number = 1; number < 200; number++) {
int quant = 0;
for (int divisor = 1; divisor <= number; divisor++)
if (number % divisor == 0)
quant++;
printf ("%3d: (%2d):", number, quant);
quantN[quant]++;
for (int divisor = 1; divisor <= number; divisor++)
if (number % divisor == 0)
printf (" %3d", divisor);
putchar ('\n');
}
putchar ('\n');
for (int quant = 0; quant < 200; quant++)
if (quantN[quant] > 0)
printf ("Quantity with %2d factors is %d\n",
quant, quantN[quant]);
return 0;
}
The relevant lines of that output are:
60: (12): 1 2 3 4 5 6 10 12 15 20 30 60
72: (12): 1 2 3 4 6 8 9 12 18 24 36 72
84: (12): 1 2 3 4 6 7 12 14 21 28 42 84
90: (12): 1 2 3 5 6 9 10 15 18 30 45 90
96: (12): 1 2 3 4 6 8 12 16 24 32 48 96
108: (12): 1 2 3 4 6 9 12 18 27 36 54 108
126: (12): 1 2 3 6 7 9 14 18 21 42 63 126
132: (12): 1 2 3 4 6 11 12 22 33 44 66 132
140: (12): 1 2 4 5 7 10 14 20 28 35 70 140
150: (12): 1 2 3 5 6 10 15 25 30 50 75 150
156: (12): 1 2 3 4 6 12 13 26 39 52 78 156
160: (12): 1 2 4 5 8 10 16 20 32 40 80 160
198: (12): 1 2 3 6 9 11 18 22 33 66 99 198
Quantity with 12 factors is 13
The complete output of that is shown below, and I'll ask in advance for forgiveness for butchering the English language with the phrase "Quantity with 1 factors is 1" but I don't see the need for making the program grammatically aware for what is, in essence, a one-off activity.
Or you can just take that to mean I'm basically lazy :-)
1: ( 1): 1
2: ( 2): 1 2
3: ( 2): 1 3
4: ( 3): 1 2 4
5: ( 2): 1 5
6: ( 4): 1 2 3 6
7: ( 2): 1 7
8: ( 4): 1 2 4 8
9: ( 3): 1 3 9
10: ( 4): 1 2 5 10
11: ( 2): 1 11
12: ( 6): 1 2 3 4 6 12
13: ( 2): 1 13
14: ( 4): 1 2 7 14
15: ( 4): 1 3 5 15
16: ( 5): 1 2 4 8 16
17: ( 2): 1 17
18: ( 6): 1 2 3 6 9 18
19: ( 2): 1 19
20: ( 6): 1 2 4 5 10 20
21: ( 4): 1 3 7 21
22: ( 4): 1 2 11 22
23: ( 2): 1 23
24: ( 8): 1 2 3 4 6 8 12 24
25: ( 3): 1 5 25
26: ( 4): 1 2 13 26
27: ( 4): 1 3 9 27
28: ( 6): 1 2 4 7 14 28
29: ( 2): 1 29
30: ( 8): 1 2 3 5 6 10 15 30
31: ( 2): 1 31
32: ( 6): 1 2 4 8 16 32
33: ( 4): 1 3 11 33
34: ( 4): 1 2 17 34
35: ( 4): 1 5 7 35
36: ( 9): 1 2 3 4 6 9 12 18 36
37: ( 2): 1 37
38: ( 4): 1 2 19 38
39: ( 4): 1 3 13 39
40: ( 8): 1 2 4 5 8 10 20 40
41: ( 2): 1 41
42: ( 8): 1 2 3 6 7 14 21 42
43: ( 2): 1 43
44: ( 6): 1 2 4 11 22 44
45: ( 6): 1 3 5 9 15 45
46: ( 4): 1 2 23 46
47: ( 2): 1 47
48: (10): 1 2 3 4 6 8 12 16 24 48
49: ( 3): 1 7 49
50: ( 6): 1 2 5 10 25 50
51: ( 4): 1 3 17 51
52: ( 6): 1 2 4 13 26 52
53: ( 2): 1 53
54: ( 8): 1 2 3 6 9 18 27 54
55: ( 4): 1 5 11 55
56: ( 8): 1 2 4 7 8 14 28 56
57: ( 4): 1 3 19 57
58: ( 4): 1 2 29 58
59: ( 2): 1 59
60: (12): 1 2 3 4 5 6 10 12 15 20 30 60
61: ( 2): 1 61
62: ( 4): 1 2 31 62
63: ( 6): 1 3 7 9 21 63
64: ( 7): 1 2 4 8 16 32 64
65: ( 4): 1 5 13 65
66: ( 8): 1 2 3 6 11 22 33 66
67: ( 2): 1 67
68: ( 6): 1 2 4 17 34 68
69: ( 4): 1 3 23 69
70: ( 8): 1 2 5 7 10 14 35 70
71: ( 2): 1 71
72: (12): 1 2 3 4 6 8 9 12 18 24 36 72
73: ( 2): 1 73
74: ( 4): 1 2 37 74
75: ( 6): 1 3 5 15 25 75
76: ( 6): 1 2 4 19 38 76
77: ( 4): 1 7 11 77
78: ( 8): 1 2 3 6 13 26 39 78
79: ( 2): 1 79
80: (10): 1 2 4 5 8 10 16 20 40 80
81: ( 5): 1 3 9 27 81
82: ( 4): 1 2 41 82
83: ( 2): 1 83
84: (12): 1 2 3 4 6 7 12 14 21 28 42 84
85: ( 4): 1 5 17 85
86: ( 4): 1 2 43 86
87: ( 4): 1 3 29 87
88: ( 8): 1 2 4 8 11 22 44 88
89: ( 2): 1 89
90: (12): 1 2 3 5 6 9 10 15 18 30 45 90
91: ( 4): 1 7 13 91
92: ( 6): 1 2 4 23 46 92
93: ( 4): 1 3 31 93
94: ( 4): 1 2 47 94
95: ( 4): 1 5 19 95
96: (12): 1 2 3 4 6 8 12 16 24 32 48 96
97: ( 2): 1 97
98: ( 6): 1 2 7 14 49 98
99: ( 6): 1 3 9 11 33 99
100: ( 9): 1 2 4 5 10 20 25 50 100
101: ( 2): 1 101
102: ( 8): 1 2 3 6 17 34 51 102
103: ( 2): 1 103
104: ( 8): 1 2 4 8 13 26 52 104
105: ( 8): 1 3 5 7 15 21 35 105
106: ( 4): 1 2 53 106
107: ( 2): 1 107
108: (12): 1 2 3 4 6 9 12 18 27 36 54 108
109: ( 2): 1 109
110: ( 8): 1 2 5 10 11 22 55 110
111: ( 4): 1 3 37 111
112: (10): 1 2 4 7 8 14 16 28 56 112
113: ( 2): 1 113
114: ( 8): 1 2 3 6 19 38 57 114
115: ( 4): 1 5 23 115
116: ( 6): 1 2 4 29 58 116
117: ( 6): 1 3 9 13 39 117
118: ( 4): 1 2 59 118
119: ( 4): 1 7 17 119
120: (16): 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
121: ( 3): 1 11 121
122: ( 4): 1 2 61 122
123: ( 4): 1 3 41 123
124: ( 6): 1 2 4 31 62 124
125: ( 4): 1 5 25 125
126: (12): 1 2 3 6 7 9 14 18 21 42 63 126
127: ( 2): 1 127
128: ( 8): 1 2 4 8 16 32 64 128
129: ( 4): 1 3 43 129
130: ( 8): 1 2 5 10 13 26 65 130
131: ( 2): 1 131
132: (12): 1 2 3 4 6 11 12 22 33 44 66 132
133: ( 4): 1 7 19 133
134: ( 4): 1 2 67 134
135: ( 8): 1 3 5 9 15 27 45 135
136: ( 8): 1 2 4 8 17 34 68 136
137: ( 2): 1 137
138: ( 8): 1 2 3 6 23 46 69 138
139: ( 2): 1 139
140: (12): 1 2 4 5 7 10 14 20 28 35 70 140
141: ( 4): 1 3 47 141
142: ( 4): 1 2 71 142
143: ( 4): 1 11 13 143
144: (15): 1 2 3 4 6 8 9 12 16 18 24 36 48 72 144
145: ( 4): 1 5 29 145
146: ( 4): 1 2 73 146
147: ( 6): 1 3 7 21 49 147
148: ( 6): 1 2 4 37 74 148
149: ( 2): 1 149
150: (12): 1 2 3 5 6 10 15 25 30 50 75 150
151: ( 2): 1 151
152: ( 8): 1 2 4 8 19 38 76 152
153: ( 6): 1 3 9 17 51 153
154: ( 8): 1 2 7 11 14 22 77 154
155: ( 4): 1 5 31 155
156: (12): 1 2 3 4 6 12 13 26 39 52 78 156
157: ( 2): 1 157
158: ( 4): 1 2 79 158
159: ( 4): 1 3 53 159
160: (12): 1 2 4 5 8 10 16 20 32 40 80 160
161: ( 4): 1 7 23 161
162: (10): 1 2 3 6 9 18 27 54 81 162
163: ( 2): 1 163
164: ( 6): 1 2 4 41 82 164
165: ( 8): 1 3 5 11 15 33 55 165
166: ( 4): 1 2 83 166
167: ( 2): 1 167
168: (16): 1 2 3 4 6 7 8 12 14 21 24 28 42 56 84 168
169: ( 3): 1 13 169
170: ( 8): 1 2 5 10 17 34 85 170
171: ( 6): 1 3 9 19 57 171
172: ( 6): 1 2 4 43 86 172
173: ( 2): 1 173
174: ( 8): 1 2 3 6 29 58 87 174
175: ( 6): 1 5 7 25 35 175
176: (10): 1 2 4 8 11 16 22 44 88 176
177: ( 4): 1 3 59 177
178: ( 4): 1 2 89 178
179: ( 2): 1 179
180: (18): 1 2 3 4 5 6 9 10 12 15 18 20 30 36 45 60 90 180
181: ( 2): 1 181
182: ( 8): 1 2 7 13 14 26 91 182
183: ( 4): 1 3 61 183
184: ( 8): 1 2 4 8 23 46 92 184
185: ( 4): 1 5 37 185
186: ( 8): 1 2 3 6 31 62 93 186
187: ( 4): 1 11 17 187
188: ( 6): 1 2 4 47 94 188
189: ( 8): 1 3 7 9 21 27 63 189
190: ( 8): 1 2 5 10 19 38 95 190
191: ( 2): 1 191
192: (14): 1 2 3 4 6 8 12 16 24 32 48 64 96 192
193: ( 2): 1 193
194: ( 4): 1 2 97 194
195: ( 8): 1 3 5 13 15 39 65 195
196: ( 9): 1 2 4 7 14 28 49 98 196
197: ( 2): 1 197
198: (12): 1 2 3 6 9 11 18 22 33 66 99 198
199: ( 2): 1 199
Quantity with 1 factors is 1
Quantity with 2 factors is 46
Quantity with 3 factors is 6
Quantity with 4 factors is 59
Quantity with 5 factors is 2
Quantity with 6 factors is 27
Quantity with 7 factors is 1
Quantity with 8 factors is 31
Quantity with 9 factors is 3
Quantity with 10 factors is 5
Quantity with 12 factors is 13
Quantity with 14 factors is 1
Quantity with 15 factors is 1
Quantity with 16 factors is 2
Quantity with 18 factors is 1
Best Answer
Judging by the comments, you overlooked the one.
Here is a method that can be generalized:
Each factor of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ has the form $2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}13^{a_6}$, where $a_1, a_2, a_3, a_4, a_5, a_6 \in \{0, 1\}$. Since there are two possible choices for each of the six exponents, there are $2^6 = 64$ possible factors of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.