July 1, 1889.] 



KNOVv^LEDGE ♦ 



187 



Thus 1224, 3570, 4692=17M. 



If we proceed, as in (1) supi-a, we need not confine the 

 principle to figures of two digits. Thus 53 x 2 = 10(5. 

 Therefore .^i406=17 M. Here 53 + 1=54. 

 Again, 1234x2 = 2468. Therefore 

 2468 

 1234 



125868=17M. 



If we take care to leave untouched the two figures on the 

 right, and phice the other figures, as in the example, the 

 principle will extend to any number of digits. This is 

 easily proved thus : — Let t signify ten. 



Then (a< + i)(<^ + 2) = 17M, 



i.e., al^ + bf^ + 2at + 2h^7M. Q.E.D. 



Also, since {at^ + bt + c){t^ + 'i) =: 17 K, 



The-efore at^' + bt' -\- cfi + 3at- + 3bt + 3c= 17M. 

 This means that if you multiply a number of three digits 

 by 3 and place the six figures .side by side, the number is 

 then divisible by 17. 



(4) Similarly, if we take any number of three digits and 

 double it, the figures placed side by side are exactly divisible 

 by 167. Thus 123x2 = 246. Then 123246=167 M. 

 Similarly 234x2=468. Thus 234468=167M. Nor need 

 we confine the number to thiee digits, if we ]n-oceed on the 

 principle given above. Thus 0789x2^13578. Therefore 

 6802578=167M, since 6789 + 13=6802, Ac. The proof is 

 similar to that of No. 3, and the principle is of wide 

 application. £!.g., we may take any 5 figures and double 

 them : the ten figures are divisible by 7. Thus 



l66293725S=7M. 



(5) A general principle applicable to all numbers ending 

 in 1, 3, 7, or 9 is as follows * : — 



Take any number, .say, 123456789 = 13R + 1. Then 

 1123456789 or 12123456787, 

 2123456790 or 111234567fc6, 

 3123456791 or 10123456785, 

 (fee. &c. &c. ad injin. 



■ure all divisible by 13, without remainder. 

 Again, take 12349 = 7M + 1. 

 Then 412349 or 312347, 



8123.50 or 612346, 

 1212351 or 912345, 

 1612352 or 1212344, 

 etc. itc. ad inJin. 



are all exactly divisible by 7. 



It will be readily perceived that the prefixes 4, 8, 12, 

 16, &c., also 3, 6, 9, 12, &:c., are in arithmetical pro- 

 gression, as also the figures in the unit's place. The prin- 

 ciple may be briefly stated in mathematical language as 

 follows : — 



np 10"' + N{M) + n=N(R) 

 where »V and n are any integers ; m the number of digits, p 

 the prefix. The proof is not diilicult. We know that 21, 

 301, 1001, 50001, 400001, 6000001, are divisible by 7. 

 These prefixes 2, 3, 1, 5, 4, 6 follow the law Sjo^ — 7.'-= 

 •Pd+i- Thus 5 X 2 — 7 = 3, &c. ; and these values recur. 

 Thus if N=»(R) + 1 by hypothesis say 22 = 7(R) + 1, 

 And we know that ;^«"' + l=/((R,) say 301 = 7(R|), 

 Then ;;«'" + N=h(R.,) or 3-22 = 7(R2), 



Also 2;)<"' + N + l=«(R3) or 6-23 = 7(R.,); 



Also 3jo<'" + N + 2=»(R,) or 9-24 = 7(R,'). 



Thus npt"" + N + /; — 1 =«(R,), 

 or, since N=?i(R) + l, 



n;)10"' + N(R) + >i=N(R). 

 Similarly for .any prime niuubor, say 19, the prefixes are 



* Multiples of these numbers only can have all the digits in the 

 'units place. Even numbers cannot nor those ending in 5. 



17, 15, 11, 3, 6, 12, 5, 10, 1, according to the law 



19, — 2", and the values recur after -^^ values. 



2 



(6) It is not difficult to get the prefixes without actual 

 trial or division, e.rj., What prefix will make the number 



123456789 exactly divisible by 7 1 

 Our formula for seven is 7a;— 3"?-, i.e. 7x4 — 3' x 1 = 1. 

 Therefore 1123456789 is exactly divisible by 7. 



Again, what prefix will make 123456789 exactly 

 divisible by 13? Our formula for 13 is 13./: — 4°r, i.e. 

 13x5-4''xl = l. Therefore 1123456789 is also exactly 

 divisible by 13. What prefi.x; will make 1236 divisible by 

 19? Using our formula for 19, viz. 19a:— 2"r we get 

 19-2°=3. Thus 31236, 61237, 91238, &c., &c., ad in^n. 

 are divisible by 19. 



A few useful formulse are — 



2°10"-1 = 19R. 

 3„ 106">-"_1 = 7R 

 4° 10*'"+"-l = 1.5R. 



(7) No. (5) supra enables us to obtain a convenient test 

 for divisibility by any prime 75, e.g. : — 



If 123456*3= 19M, then 123456 = 19M +13. 



Therefore 123443=19M, and 12344=19Mo + 13. 

 Then 12331 = 19M3 and 1233 = 19M^" + 17. 



Thus 1216 = 19Mj and 121 = 19.\I,+ 7. 



Therefore 114 = 19M,, etc. 

 The proof is simple. We have 



19x9:=171 19x4=76 



19x8=152 19x3=57 



19x7 = 133 19x2=38 



19x6 = 114 19x1 = 19 



19x5= 95 

 If a number then end in any digit, say 3, and is also 

 divisible by 19, then the other figures (however numerous) 

 are of the Form 19M + 1 3. 



We see also from the table that if we multiply any of the 

 digits by 17 and reject 19.c, we get the figures on the left. 



A similar law applies to any of the primes. Hence the 

 rules. 



Let us test the number 123454 for 17. 

 Since 17 x3 = 51 our multiplier for 17 is 5. 

 Say, then, 5 x4=20, reject 17, leaving 3 and 3 from the 

 next figure 5=2, thus 12342=1751,. 



Proceeding, say 5x2=10 and subtract, lewing 1224= 

 17M,. 



Proceeding again, say 5x4 = 20, reject 17 = 3 and take 

 3 for 122 = 119=17M3. 



Since the last result is true, therefore 123454 is a multiple 

 by 17. 



After a little practice the process can be gone through 

 rapidly. 



(8) It is generally known that we may square any 

 number ending in 5 as follows : — Square 65. Here 

 5x5=25; 6x7=42. Thus 65-=4225. But it is not 

 generally known, as can be ejisily proved mathematically, 

 that the same principle applies to anj' two numbers whose 

 tens' digits are alike, and whose units equal 10, eg. 

 multiply 86 by 84. Here 8 x 9=72 ; 4 x 6=24. 



Therefore our result is 7224. 



Again, multiply 177 by 173. Here 17x18=306; 

 7 x 3=21. Therefore the result is 30621. 



Messrs. Hurray have sent us a new and cheap edition of 

 Darwin's " Voyage of the Beagl-;," the copyright in which 

 appears now to have run out. Probably no book has done 

 more to make naturalists than this simple narration of 

 Darwin's earlier travels .and observations. 



