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

• 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
[...]