Jdly 4, 1884.] 



- KNOWLEDGE • 



15 



auJiblc actions. . . When we wish to express a hitherto unnamed 

 colour, the simplest way of dointj it is to taue an object which 

 possesses that colour and apply its title as an adjective to the 

 thing which we wish to describe. A particular shade of very light 

 yellow has no distinctive name, but we must call it something for 

 some special purpose, and so we thiuk of its nearest common 

 representative, a primrose. Thenceforward the new name becomes 

 an adjective, and we ask naturally for a yard of primrose ribbon. 

 Now, what we see civilised men doing to-day under our own eyes, 

 primitive men did centuries ago, when they framed the earliest 

 colour names. It would seem at present as though the various 

 terms for colours might be divided into two classes — the truly 

 abstract, such as blue, green, yellow, and the concrete used abstractly, 

 such as lilac, orange, pinlc. The former class a]>pears to have no 

 other meaning than that of pure colour; while the latter class are 

 clearly derived from the names of concrete objects. But in reality 

 the difference between them is merely one of time. Abstract 

 colour-terms are the names of concretes, whose original significa- 

 tion has been forgotten" (pp. 252, 253). Edward Clodd. 



DIVISIBILITY BY SEVEN. 



8 

 931 



P24 



026 



oys 



518 I 

 423 



-123 



[1324] — Mr. Askew, in 1274, May 30, asks for a proof of a 

 method for ascertaining the divisibility of a numbei" by 7, which he 

 states to have been discovered by Mr. Rickard, of Birmingham. 

 Probably many have discovered it : my father did. for one, and 

 taught it to me some thirty years ago. The test -number is equally 

 xiseful for 7, 11, and 13. The method, as worked liy my father, 

 gives, in the case of a number divisible by all three factors, the 

 other factor as well, without further labour: and in this respect it 

 has an advantage over that of Mr. Bickard. 



If a number, X, be marked off from the right-hand end in periods 

 of three digits ; and if a, b, c, &c., be the periods ; and if M be the 

 difference between the sums of the alternate periods; we have, 

 writing r for 1000, 



Ts=a + br + cr' + dr^ + &c. 



M = a— i)-^c — d + &c. 



.•.N-M = b {r + l)+c(r'-l) t- <£()■» -i- 1) -H &c. 



and is divisible by (r-fl)j hence, if M be divisible by (r-H) or 



anv factor of it, so also is N. And in this case r + 1 = 1001 = 7 x 11 



xi3. 



My father's rule was to set the right-hand period under the next, 

 and subtract, setting the remainder under 

 the next, and so on. In the last period, the 

 subtraction is doicnwards if the lower number 

 be the larger. In this instance, since we 

 have 1 to carry into the last period, the 931 

 must be read as 932. The ultimate remainder, 

 024, is the test-nnmber ; and, since this is divisible by 7 and 11, so 

 also is the whole number. 



If the test-number chanced to be zero, the second line would be 

 the cpiotient produced by dividing the given number by 1001 ; i.e., it 

 is the factor remaining after dividing out 7, 11, and 13. For let us 

 call the second line " V; " writing three ciphers at the end, we get 

 lOOOV ; and we know that, if this be deducted from the upper line, 

 the remainder = V. Hence N = lOOlV = 7 x 11 x 13 x V. In 

 the above example, if the left-hand period were 932 instead of 8, 

 the test-number would be zero. 



If the periods be single digits, i.e., if 7- = 10, we get a test for 

 divisibility by 11, and at the same time the 



G4372583 quotient after dividing out 11. The rule is to 



05852053 set the last digit under the next, and subtract, 



setting the remainder under the next, and so on. 



In this instance the test-number =0; hence the given number 



= 11x5852053. 



With periods of two digits, we get a test for divisibility by 101 ; 

 and so for four or more digits. C. L. Dodgsox. 



Ch. Ch., Oxford. 



P.S. — The sum of all the periods gives us, for periods of 1, 2, 3, 

 &c., digits, a test for divisibility by 9, 99, 999 ( = 27 x 37), &c., or 

 for any factors of these numbers. This method may also be 

 worked by a rule analogous to that given above ; e.;;., to test for 

 999, mark off in periods of three, write 000 over the right-hand 

 period, and subtract, writing the remainder over the next, and so 

 on. Hence, also, if the test-number chanced to be zero, the upper 

 line (omitting the 000) would be the quotient produced by dividing 

 the given number by 999. 



Probably similar rules may be made for most primes. I have 

 myself made fairly simple rules for 17 and for 19; but such pro- 

 cesses are rather ctirions than useful. 



[1325] — Mr. Askew's rule is applicable not only, as he says, to 

 the division 7 and 13, but also to 91. The reason is that both 



(1000 -t-l) and (1000--1) happen to be exactly multiples of these 

 three numbers. 



Any number of over six figures may be expressed in the form 



a -I- lOOOt. -f lOOOV -H lOOO^a -f [A] 



where the letters a, b, &c., stand for the numbers expressed by the 

 successive triplets of figures, beginning at the right hand. 

 Now the quantitv 



(10UO-fl)!--t-(IO0O'-l)c-h(1000=-i-l)d-(- [B] 



must be a multiple of 7, 13, and 91. If therefore the quantity [A] 

 be divisible by one of these numbers, so also will the difference 



between [A] and [B] — viz, a — b + c-d+ 



Henry Bradley. 



[A number which is divisible by two prime numbers must neces- 

 sarily be divisible by their product. — K.P.j. 



(Answer to Letter 1274, page 399.) 



[1326] — " Point off the given number into periods of three 

 figures. Add separately alternate periods, and find the difference 

 of the sums thus obtained. If this difference is divisible by seven, 

 the original number is so, and if not, not." 



The reason is as follows : — Adding the figures alternately, and 

 taking the difference, gives the criterion of divisibility by eleven, 

 since the successive powers of ten (1, 10, 100, 1,000, &c.) divided 

 by eleven give remainders 1, 10, 1, 10, &c. ; similarly, adding them 

 alternatelv in sets of two gives the criterion of divisibility by 101, 

 since 1, I'OO, 10,000, 1,000,000, &c., divided by 101, give as re- 

 mainders 1, 100, 1, 100, &c., and adding alternately in sets of three 

 gives the criterion of divisibility by 1001, since 1, Iff", lO", 10', ic, 

 give the remainders 1, 10', 1, 10', &c., when divided by 1001, and 

 since 1001 is divisible by 7 and 13, the same method gives a 

 criterion of divisibility for each of those numbers. 



This is only a special case of the law of divisibility in any scale 

 of notation— viz., that adding the figures gives the criterion of 

 divisibility by one les.-i than the radix of the scale, adding them 

 alternately and finding the difference gives the criterion for one 

 more than the radix. The radix in this case is 1,000. The number 

 taken by Mr. Askew is 220,97-1,901. 



220,000,000 = a multiple of 1,001 + 220 



974,000 = „ ,, - 974 



901 = „ „ + 901 



.•. 220,974,901 = „ ,, + 147 



and since 1,001 is divisible by 7 and 147 is divisible by 7, a multiple 

 of 1,001 -rl47 is divible by 7. Also since 1,001 is divisible by 13 

 and 147 divided by 13 leaves remainder 4, a multiple of 1,001 + 147 

 must leave remainder 4. H. A. Nesbitt. 



PKOPEP>TY OF NUMBERS. 



[1327]— The problem proposed in article No. 1295 of your 

 issue ©f 6th inst. presents no great difiiculty. The actual working- 

 out, which is rather lengthy, I forward separately. 



The result may be given as follows : — 



Let a, b, c, and d be the remainders when a number N is 

 divided by 3, 5, 7, and 11 respectively. 



Then N = lloSr -h 3S5a -H 231t) -f 330"c - 210d. 



The value of r must be taken to suit the conditions of the pro- 

 blem, viz., N is to be a number of three digits. 



If this problem is to be given as a mode of figure-conjuring, the 

 conjurer will find the figures involved rather long for mental calcu- 

 lation. It would be better in that case to limit the problem thus :— - 



Tell a person to take any number not exceeding 100 (to say 105 

 might give a clue to the puzzle), to mentally divide the number by 

 3, bv 5, and by 7, and to announce the remainders. The would-be 

 conjurer should then be able to announce the number thought of in 

 a few seconds. 



The kev to the conjuring is as follows — 



S = i05r + 70a + 2lb-rloc 

 r being taken as zero or a negative quantity to suit conditions of 

 problem. A- H- B. E. 



P.S. — If another divisor, 13, be introduced and e be the remainder, 

 then it can similarly be found that 



N = 15015)- f 5005a -i- 0006b + 10725c + \3God + -6930?. 



PROBLEM IX XDMBERS. 



[1328]— The following problem was given by a Moor at Gibraltar 

 to a friend of mine ; — 



Divide any number by 3, 5, and 7 ; the remainders are a, b, and c, 

 and the number of 100 "is d. With these data what is the number ? 



