Root Pursuit

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:

• √2 = 1·414213562373095048801688724209698078569671875376948073...

It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:

• 2 = 2^1
• 4 = 2^2
• 8 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
[...]

But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?

131072 = 2^17
129140163 = 3^17
1255420347077336152767157884641... = 2^193
1214512980685298442335534165687... = 3^193
2175541218577478036232553294038... = 2^619
2177993962169082260270654106078... = 3^619
7524389324549354450012295667238... = 2^2016
7524012611682575322123383229826... = 3^2016

There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:

32 = 2^5
3125 = 5^5
316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185

The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:

• √10 = 3·1622776601683793319988935444327185337195551393252168268575...

I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:

• 316227... = 2^2728361
• 316227... = 5^2728361
• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 3162277660... = 2^961700165
• 3162277660... = 5^961700165

But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:

32 = 2^5
3125 = 5^5
• 3^2 = 9

316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
• 31^2 = 961

3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
• 3162^2 = 9998244

3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185
• 31622^2 = 999950884

• 316227... = 2^2728361
• 316227... = 5^2728361
• 316227^2 = 99999515529

• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 3162277^2 = 9999995824729

• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 31622776^2 = 999999961946176

• 3162277660... = 2^961700165
• 3162277660... = 5^961700165
• 3162277660^2 = 9999999998935075600

The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:

3 162266908803418110961625404267... = 2^127185
3·162277660168379331998893544432... = √10
3 162288411569894029343799063611... = 5^127185

In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:

• 2.24103122055214532500432040411... = √6 (in base 6)

24 = 2^4
213 = 3^4
225522024 = 2^34 in base 6 = 2^22 in base 10
22225525003213 = 3^34 (3^22)
2241525132535231233233555114533... = 2^1303 (2^327)
2240133444421105112410441102423... = 3^1303 (3^327)
2241055222343212030022044325420... = 2^153251 (2^15007)
2241003215453455515322105001310... = 3^153251 (3^15007)
2241032233315203525544525150530... = 2^233204 (2^20164)
2241030204225410320250422435321... = 3^233204 (3^20164)
2241031334114245140003252435303... = 2^2110415 (2^102539)
2241031103430053425141014505442... = 3^2110415 (3^102539)

And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:

• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

55AA4 = 2^M in base 30 = 2^22 in base 10
5NO6CQN69C3Q0E1Q7F = F^M = 15^22
5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606)
5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606)
5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934)
5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934)
[...]
5R4C9 = 3^E in base 30 = 3^14 in base 10
52CE6A3L3A = A^E = 10^14
5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113)
5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113)
5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187)
5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187)
[...]
5OCNCNRAP = 5^I in base 30 = 5^18 in base 10
54NO22GI76 = 6^I (6^18)
5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567)
5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567)
5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786)
5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786)
[...]

So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:

•  √2 = 1·414213562373095048801688724209698078569671875376948073...
• √20 = 4·472135954999579392818347337462552470881236719223051448...

16 = 2^4
125 = 5^3
140737488355328 = 2^47
142108547152020037174224853515625 = 5^46
1413... = 2^243
1414... = 5^242
14141... = 2^6651
14142... = 5^6650
141421... = 2^35389
141420... = 5^35388
4472136... = 2^162574
4472135... = 5^162573
141421359... = 2^3216082
141421352... = 5^3216081
447213595... = 2^172530387
447213595... = 5^172530386
[...]

Shareway to Seven

An adaptation of an interesting distribution puzzle from Joseph Degrazia’s Math is Fun (1954):

After a successful year of plunder on the high seas, a pirate ship returns to its island base. The pirate chief, who enjoys practical jokes and has a mathematical bent, hands out heavy bags of gold coins to his seven lieutenants. But when the seven lieutenants open the bags, they discover that each of them has received a different number of coins.

They ask the captain why they don’t have equal shares. The pirate chief laughs and tells them to re-distribute the coins according to the following rule: “At each stage, the lieutenant with most coins must give each of his comrades as many coins as that comrade already possesses.”

The lieutenants follow the rule and each one in turn becomes the lieutenant with most coins. When the seventh distribution is over, all seven of them have 128 coins, the coins are fairly distributed, and the rule no longer applies.

The puzzle is this: How did the pirate captain originally allocate the coins to his lieutenants?


If you start at the beginning and work forward, you’ll have to solve a fiendishly complicated set of simultaneous equations. If you start at the end and work backwards, the puzzle will resolve itself almost like magic.

The puzzle is actually about powers of 2, because 128 = 2^7 and when each of six lieutenants receives as many coins as he already has, he doubles his number of coins. Accordingly, before the seventh and final distribution, six of the lieutenants must have had 64 coins and the seventh must have had 128 + 6 * 64 coins = 512 coins.

At the stage before that, five of the lieutenants must have had 32 coins (so that they will have 64 coins after the sixth distribution), one must have had 256 coins (so that he will have 512 coins after the sixth distribution), and one must have had 64 + 5 * 32 + 256 coins = 480 coins. And so on. This is what the solution looks like:

128, 128, 128, 128, 128, 128, 128
512, 64, 64, 64, 64, 64, 64
256, 480, 32, 32, 32, 32, 32
128, 240, 464, 16, 16, 16, 16
64, 120, 232, 456, 8, 8, 8
32, 60, 116, 228, 452, 4, 4
16, 30, 58, 114, 226, 450, 2
8, 15, 29, 57, 113, 225, 449

So the pirate captain must have originally allocated the coins like this: 8, 15, 29, 57, 113, 225, 449 (note how 8 * 2 – 1 = 15, 15 * 2 – 1 = 29, 29 * 2 – 1 = 57…).

The puzzle can be adapted to other powers. Suppose the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades twice as many coins as that comrade already possesses.” If the pirate captain has six lieutenants, after each distribution each of five will have n + 2n = three times the number of coins that he previously possessed. The six lieutenants each end up with 729 coins = 3^6 coins and the solution looks like this:

13, 37, 109, 325, 973, 2917
39, 111, 327, 975, 2919, 3
117, 333, 981, 2925, 9, 9
351, 999, 2943, 27, 27, 27
1053, 2997, 81, 81, 81, 81
3159, 243, 243, 243, 243, 243
729, 729, 729, 729, 729, 729

For powers of 4, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades three times as many coins as that comrade already possesses.” With five lieutenants, each of them ends up with 1024 coins = 4^5 coins and the solution looks like this:

16, 61, 241, 961, 3841
64, 244, 964, 3844, 4
256, 976, 3856, 16, 16
1024, 3904, 64, 64, 64
4096, 256, 256, 256, 256
1024, 1024, 1024, 1024, 1024

For powers of 5, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades four times as many coins as that comrade already possesses.” With four lieutenants, each of them ends up with 625 coins = 5^4 coins and the solution looks like this:

17, 81, 401, 2001
85, 405, 2005, 5
425, 2025, 25, 25
2125, 125, 125, 125
625, 625, 625, 625