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For how many integers where \(2 \le n \le 100\) is \(\binom{n}{2}\) odd?

 Oct 10, 2020
edited by WarpPrime  Oct 10, 2020
 #1
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n

2 = 1
3 = 3
6 = 15
7 = 21
10 = 45
11 = 55
14 = 91
15 = 105
18 = 153
19 = 171
22 = 231
23 = 253
26 = 325
27 = 351
30 = 435
31 = 465
34 = 561
35 = 595
38 = 703
39 = 741
42 = 861
43 = 903
46 = 1035
47 = 1081
50 = 1225
51 = 1275
54 = 1431
55 = 1485
58 = 1653
59 = 1711
62 = 1891
63 = 1953
66 = 2145
67 = 2211
70 = 2415
71 = 2485
74 = 2701
75 = 2775
78 = 3003
79 = 3081
82 = 3321
83 = 3403
86 = 3655
87 = 3741
90 = 4005
91 = 4095
94 = 4371
95 = 4465
98 = 4753
99 = 4851
Total odds = 50

 Oct 10, 2020
 #2
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Yes, that is right, but is there a simpler way to do it, Guest?

WarpPrime  Oct 10, 2020
 #3
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If you look at it carefully, you will see that from 6 and up every  FOURTH and FIFTH numbers are odd. So:

6 + 4=10 and 6 + 5 =11  and 10 + 4=14  and 10 + 5=15  and 14 +4=18 and 14 + 5 =19.....and so on.

Guest Oct 10, 2020

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