2 The pattern of the positive power end number

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:

7. ...

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