2 positive Powers-2, 4, 8, 16, 32, 64, 128, 256, ...-at the end of the numbers follow an obvious rule: 2, 4, 8, 6, 2, 4, 8, 6, .... These 4 numbers are always going to loop. In addition to the last number there is a loop-actually the final m bit-a power of 2 starting at 2m. For example, there is a loop of length 20 for the last two digits starting at 04, and a loop of length 100 for the last 3 digits starting from 008.
In this article, I will show you why there are these loops, how long they are, how they are expressed in mathematical form, and how they are seen.
The loop at the end number
The bottom digit-the position-is the remainder of the decimal integer d after d/10. Equivalently, the end number is the result of the D mod 10, with the minimum non-negative-normal remainder-as the result of the Convention. Modulo operation, which is continuously superimposed to the power of 2, to get the last number of loops:
- ...
We start from 2, take 10 modulo, multiply by 2, then 10 modulo, and so on. This pattern will loop until a previous result-2 appears again in the fifth step-loop confirmation. This shows the number 2n, n≥1, the lowest number in four digits 2, 4, 8, and 6 cycles.
This cycle tells us that the power of the same 2 of the end number is correlated, and their exponent differs by 4:
- Ends in 2: 21, 25, 29, 213, 217, ....
- Ends in 4: 22, 26, 210, 214, 218, ....
- Ends in 8: 23, 27, 211, 215, 219, ....
- Ends in 6: 24, 28, 212, 216, 220, ....
You can use the exponential arithmetic rules to describe more succinctly, showing the top 4 of all the last numbers in the power of 2:
- Ends in 2: 21 24k, or 21+4k, k≥0.
- Ends in 4: 22 24k, or 22+4k, k≥0.
- Ends in 8: 23 24k, or 23+4k, k≥0.
- Ends in 6: 24 24k, or 24+4k, k≥0.
It is also possible to establish a connection to a power of 2 according to the results of the exponential 4 modulo:
- Ends in 2:
- Ends in 4:
- Ends in 8:
- Ends in 6:
It is easy to know that the number of the positive power of any 2 is a few. For example, the lowest number of 2319 is 8, because.
Last
Cycle in the last Digit of 2n, n≥1
Power of both (k≥0) |
Exponent (mod 4) | Digit |
21+4k |
1 |
2 |
22+4k |
2 |
4 |
23+4k |
3 |
8 |
24+4k |
0 |
6 |
Summary, the form tells us, if,.
The last two digits of the loop
A similar analysis, just for 100 modulo, shows the last two digits of the power of 2, starting from the beginning, the cycle time is:
Last
Cycle in the last of Digits of 2n, n≥2
Power of both (k≥0) |
Exponent (MoD) | 2 Digits |
22+20k |
2 |
04 |
23+20k |
3 |
08 |
24+20k |
4 |
16 |
25+20k |
5 |
32 |
26+20k |
6 |
64 |
27+20k |
7 |
28 |
28+20k |
8 |
56 |
29+20k |
9 |
12 |
210+20k |
10 |
24 |
211+20k |
11 |
48 |
212+20k |
12 |
96 |
213+20k |
13 |
92 |
214+20k |
14 |
84 |
215+20k |
15 |
68 |
216+20k |
16 |
36 |
217+20k |
17 |
72 |
218+20k |
18 |
44 |
219+20k |
19 |
88 |
220+20k |
0 |
76 |
221+20k |
1 |
52 |
The last three-bit loop
In order to find the last three digits of the cycle, repeat the above process, 1000 modulo. The following shows the last three bits of the power of 2 , starting at the beginning of the cycle of 100:
Last
Cycle in the last three Digits of 2n, n≥3
Power of both (k≥0) |
Exponent (MoD) | 3 Digits |
23+100k |
3 |
008 |
24+100k |
4 |
11W |
25+100k |
5 |
032 |
26+100k |
6 |
064 |
27+100k |
7 |
128 |
28+100k |
8 |
256 |
29+100k |
9 |
512 |
210+100k |
10 |
024 |
211+100k |
11 |
048 |
212+100k |
12 |
096 |
213+100k |
13 |
192 |
214+100k |
14 |
384 |
215+100k |
15 |
768 |
216+100k |
16 |
536 |
217+100k |
17 |
072 |
218+100k |
18 |
144 |
219+100k |
19 |
288 |
220+100k |
20 |
576 |
221+100k |
21st |
152 |
222+100k |
22 |
304 |
223+100k |
23 |
608 |
224+100k |
24 |
216 |
225+100k |
25 |
432 |
226+100k |
26 |
864 |
227+100k |
27 |
728 |
228+100k |
28 |
456 |
229+100k |
29 |
912 |
230+100k |
30 |
824 |
231+100k |
31 |
34] |
232+100k |
32 |
296 |
233+100k |
33 |
592 |
234+100k |
34 |
184 |
235+100k |
35 |
368 |
236+100k |
36 |
736 |
237+100k |
37 |
472 |
238+100k |
38 |
944 |
239+100k |
39 |
888 |
240+100k |
40 |
776 |
241+100k |
41 |
552 |
242+100k |
42 |
104 |
243+100k |
43 |
208 |
244+100k |
44 |
416 |
245+100k |
45 |
832 |
246+100k |
46 |
664 |
247+100k |
47 |
328 |
248+100k |
48 |
656 |
249+100k |
49 |
312 |
250+100k |
50 |
624 |
251+100k |
51 |
248 |
252+100k |
52 |
496 |
253+100k |
53 |
992 |
254+100k |
54 |
984 |
255+100k |
55 |
968 |
256+100k |
56 |
936 |
257+100k |
57 |
872 |
258+100k |
58 |
744 |
259+100k |
59 |
488 |
260+100k |
60 |
976 |
261+100k |
61 |
952 |
262+100k |
62 |
904 |
263+100k |
63 |
808 |
264+100k |
64 |
616 |
265+100k |
65 |
232 |
266+100k |
66 |
464 |
267+100k |
67 |
928 |
268+100k |
68 |
856 |
269+100k |
69 |
712 |
270+100k |
70 |
424 |
271+100k |
71 |
848 |
272+100k |
72 |
696 |
273+100k |
73 |
69W |
274+100k |
74 |
784 |
275+100k |
75 |
568 |
276+100k |
76 |
136 |
277+100k |
77 |
272 |
278+100k |
78 |
544 |
279+100k |
79 |
088 |
280+100k |
80 |
176 |
281+100k |
81 |
65W |
282+100k |
82 |
704 |
283+100k |
83 |
408 |
284+100k |
84 |
816 |
285+100k |
85 |
632 |
286+100k |
86 |
264 |
287+100k |
87 |
528 |
288+100k |
88 |
15W |
289+100k |
89 |
112 |
290+100k |
90 |
224 |
291+100k |
91 |
448 |
292+100k |
92 |
896 |
293+100k |
93 |
792 |
294+100k |
94 |
584 |
295+100k |
95 |
168 |
296+100k |
96 |
336 |
297+100k |
97 |
672 |
298+100k |
98 |
344 |
299+100k |
99 |
688 |
2100+100k |
0 |
376 |
2101+100k |
1 |
752 |
2102+100k |
2 |
504 |
End m-Number loop
2 of the positive power of the end of the M-bit cycle to the 10m, the cycle is 4 5m-1, starting at 2m. (The specific proof relates to number theory, beyond the scope of this article)
with
Cycle Length for number of ending Digits (1 to ten)
m |
Period (4 5m-1) |
starts |
1 |
4 |
21st |
2 |
20 |
22 |
3 |
100 |
23 |
4 |
500 |
24 |
5 |
2500 |
25 |
6 |
12500 |
26 |
7 |
62500 |
27 |
8 |
312500 |
28 |
9 |
1562500 |
29 |
10 |
7812500 |
210 |
Cycle growth is fast-exponential growth-so a list of m greater than 3 cannot be listed.
Looping nesting (Nesting of Cycles)
For the end m bit, m-1 bit, m-2 bit, ..., the last 1 bits of the loop, can be considered nested, although their starting point is staggered. You can align the smaller starting numbers by just 0.
For example, in the last three-bit loop with a length of 100, 5 times the length is 20 of the last two bits of the loop, each length is 20 of the last two-bit loop contains 5 times the last loop of length 4. You can find the lowest rule from 8 (two-bit), the last two-bit rule starts at 08 (one needs to be moved), and the last three-bit rule starts at 008 (no shift required).
The following table identifies the nested loops (all 100 rows are marked because only 100 of the 2 power-end three bits of a loop-are listed).
Nested 1-3 Digit ending Patterns from 2102
Explore the bottom loop with PARI/GP
I use PARI/GP to perform the above calculations and validations. Here are three examples:
- List the top 20 last is a power of 2 of 2 :
For (I=0,19,print ("2^", 1+4*i, ":", 2^ (1+4*i)))2^1:22^5:322^9:5122^13:81922^17:1310722^21:20971522^25: 335544322^29:5368709122^33:85899345922^37:1374389534722^41:21990232555522^45:351843720888322^49: 5629499534213122^53:90071992547409922^57:1441151880758558722^61:23058430092136939522^65:368934881474191032322^ 69:5902958103587056517122^73:94447329657392904273922^77:151115727451828646838272
- list the last two-bit loops ('% ' returns the remainder, equivalent to modulo operation):
For (I=2,21,print ("2^", I, "mod (+):", 2^i%))2^2 mod: 42^3 mod: 82^4 mod: 162^5 mod (100): 322 ^6 mod: 642^7 mod (+): 282^8 mod (+): 562^9 mod: 122^10 mod: 242^11 mod: 482^12 mod (+): 962^13 mod (+): 922^14 mod (+): 842^15 mod: 682^16 mod (+): 362^17 mod: 722^18 mod: 442^19 mod: 882^20 mod (10 0): 762^21 mod (100): 52
(Number of units here no previous complement 0)
- Print the lowest 1 to 10-digit cycle length :
for (I=1,10,print (i, ":", 4*5^ (i-1)))1:42:203:1004:5005:25006:125007:625008:3125009:156250010:7812500
2 The pattern of the positive power end number