Amusements-in-Mathematics-103

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roots	of	 the	multipliers	are	3	and	6,	or	9	and	9,	or	2	and	2,	or	5	and	8.	 I	have
divided	the	twenty-two	answers	above	into	these	four	classes.	It	is	thus	evident
that	the	digital	root	of	any	product	in	the	first	two	classes	must	be	9,	and	in	the
second	two	classes	4.
Owing	to	the	fact	that	no	number	of	five	figures	can	have	a	digital	sum	less	than
15	or	more	than	35,	we	find	that	the	figures	of	our	product	must	sum	to	either	18
or	27	to	produce	the	root	9,	and	to	either	22	or	31	to	produce	the	root	4.	There
are	3	ways	of	selecting	five	different	figures	that	add	up	to	18,	there	are	11	ways
of	 selecting	 five	 figures	 that	 add	 up	 to	 27,	 there	 are	 9	ways	 of	 selecting	 five
figures	that	add	up	to	22,	and	5	ways	of	selecting	five	figures	that	add	up	to	31.
There	are,	therefore,	28	different	groups,	and	no	more,	from	any	one	of	which	a
product	may	be	formed.
We	 next	 write	 out	 in	 a	 column	 these	 28	 sets	 of	 five	 figures,	 and	 proceed	 to
tabulate	 the	 possible	 factors,	 or	 multipliers,	 into	 which	 they	 may	 be	 split.
Roughly	speaking,	there	would	now	appear	to	be	about	2,000	possible	cases	to
be	tried,	 instead	of	the	30,240	mentioned	above;	but	the	process	of	elimination
now	begins,	 and	 if	 the	 reader	has	a	quick	eye	and	a	 clear	head	he	can	 rapidly
dispose	of	the	large	bulk	of	these	cases,	and	there	will	be	comparatively	few	test
multiplications	necessary.	It	would	take	far	 too	much	space	to	explain	my	own
method	in	detail,	but	I	will	take	the	first	set	of	figures	in	my	table	and	show	how
easily	 it	 is	 done	 by	 the	 aid	 of	 little	 tricks	 and	 dodges	 that	 should	 occur	 to
everybody	as	he	goes	along.
My	first	product	group	of	five	figures	is	84,321.	Here,	as	we	have	seen,	the	root
of	each	factor	must	be	3	or	a	multiple	of	3.	As	there	is	no	6	or	9,	the	only	single
multiplier	is	3.	Now,	the	remaining	four	figures	can	be	arranged	in	24	different
ways,	but	there	is	no	need	to	make	24	multiplications.	We	see	at	a	glance	that,	in
order	to	get	a	five-figure	product,	either	the	8	or	the	4	must	be	the	first	figure	to
the	left.	But	unless	the	2	is	preceded	on	the	right	by	the	8,	it	will	produce	when
multiplied	either	a	6	or	a	7,	which	must	not	occur.	We	are,	therefore,	reduced	at
once	 to	 the	 two	 cases,	 3	 ×	 4,128	 and	 3	 ×	 4,281,	 both	 of	 which	 give	 correct
solutions.	Suppose	next	that	we	are	trying	the	two-figure	factor,	21.	Here	we	see
that	if	the	number	to	be	multiplied	is	under	500	the	product	will	either	have	only
four	 figures	 or	 begin	 with	 10.	 Therefore	 we	 have	 only	 to	 examine	 the	 cases
21	×	843	and	21	×	834.	But	we	know	that	the	first	figure	will	be	repeated,	and
that	 the	 second	 figure	 will	 be	 twice	 the	 first	 figure	 added	 to	 the	 second.
Consequently,	as	twice	3	added	to	4	produces	a	nought	in	our	product,	the	first
case	 is	 at	 once	 rejected.	 It	 only	 remains	 to	 try	 the	 remaining	 case	 by
multiplication,	when	we	 find	 it	 does	 not	 give	 a	 correct	 answer.	 If	we	 are	 next
trying	the	factor	12,	we	see	at	the	start	that	neither	the	8	nor	the	3	can	be	in	the
units	place,	because	they	would	produce	a	6,	and	so	on.	A	sharp	eye	and	an	alert
judgment	will	enable	us	thus	to	run	through	our	table	in	a	much	shorter	time	than
would	be	expected.	The	process	took	me	a	little	more	than	three	hours.
I	have	not	attempted	to	enumerate	the	solutions	in	the	cases	of	six,	seven,	eight,
and	nine	digits,	but	I	have	recorded	nearly	fifty	examples	with	nine	digits	alone.
86.—QUEER	MULTIPLICATION.—solution
If	we	multiply	32547891	by	6,	we	get	the	product,	195287346.	In	both	cases	all
the	nine	digits	are	used	once	and	once	only.
87.—THE	NUMBER	CHECKS	PUZZLE.—solution
Divide	the	ten	checks	into	the	following	three	groups:	7	1	5—4	6—3	2	8	9	0,	and
the	first	multiplied	by	the	second	produces	the	third.
88.—DIGITAL	DIVISION.—solution
It	 is	 convenient	 to	 consider	 the	 digits	 as	 arranged	 to	 form	 fractions	 of	 the
respective	 values,	 one-half,	 one-third,	 one-fourth,	 one-fifth,	 one-sixth,	 one-
seventh,	one-eighth,	and	one-ninth.	I	will	first	give	the	eight	answers,	as	follows:
—
6729/13458	=	½
5823/17469	=	1/3
3942/15768	=	¼
2697/13485	=	1/5
2943/17658	=	1/6
2394/16758	=	1/7
3187/25496	=	1/8
6381/57429	=	1/9
The	 sum	 of	 the	 numerator	 digits	 and	 the	 denominator	 digits	 will,	 of	 course,
always	be	45,	and	the	"digital	root"	is	9.	Now,	if	we	separate	the	nine	digits	into
any	two	groups,	the	sum	of	the	two	digital	roots	will	always	be	9.	In	fact,	the	two
digital	roots	must	be	either	9—9,	8—1,	7—2,	6—3,	or	5—4.	In	the	first	case	the
actual	sum	is	18,	but	then	the	digital	root	of	this	number	is	itself	9.	The	solutions
in	the	cases	of	one-third,	one-fourth,	one-sixth,	one-seventh,	and	one-ninth	must
be	 of	 the	 form	 9—9;	 that	 is	 to	 say,	 the	 digital	 roots	 of	 both	 numerator	 and
denominator	will	be	9.	In	the	cases	of	one-half	and	one-fifth,	however,	the	digital
roots	are	6—3,	but	of	course	the	higher	root	may	occur	either	in	the	numerator	or
in	the	denominator;	thus	2697/13485,	2769/13845,	2973/14865,	3729/18645,	where,	 in	 the
first	 two	 arrangements,	 the	 roots	 of	 the	 numerator	 and	 denominator	 are
respectively	 6—3,	 and	 in	 the	 last	 two	 3—6.	 The	 most	 curious	 case	 of	 all	 is,
perhaps,	 one-eighth,	 for	 here	 the	 digital	 roots	 may	 be	 of	 any	 one	 of	 the	 five
forms	given	above.
The	denominators	of	 the	 fractions	being	 regarded	as	 the	numerators	multiplied
by	2,	3,	4,	5,	6,	7,	8,	and	9	respectively,	we	must	pay	attention	to	the	"carryings
over."	In	order	to	get	five	figures	in	the	product	there	will,	of	course,	always	be	a
carry-over	after	multiplying	 the	 last	 figure	 to	 the	 left,	and	 in	every	case	higher
than	4	we	must	carry	over	at	least	three	times.	Consequently	in	cases	from	one-
fifth	 to	 one-ninth	 we	 cannot	 produce	 different	 solutions	 by	 a	mere	 change	 of
position	 of	 pairs	 of	 figures,	 as,	 for	 example,	 we	 may	 with	 5832/17496	 and
5823/17469,	where	the	2/6	and	3/9	change	places.	It	is	true	that	the	same	figures	may
often	be	differently	arranged,	as	shown	 in	 the	 two	pairs	of	values	 for	one-fifth
that	I	have	given	in	the	last	paragraph,	but	here	it	will	be	found	there	is	a	general
readjustment	 of	 figures	 and	 not	 a	 simple	 changing	 of	 the	 positions	 of	 pairs.
There	are	other	 little	points	 that	would	occur	 to	every	solver—such	as	 that	 the
figure	5	cannot	ever	appear	to	the	extreme	right	of	the	numerator,	as	this	would
result	in	our	getting	either	a	nought	or	a	second	5	in	the	denominator.	Similarly	1
cannot	ever	appear	in	the	same	position,	nor	6	in	the	fraction	one-sixth,	nor	an
even	figure	in	the	fraction	one-fifth,	and	so	on.	The	preliminary	consideration	of
such	points	 as	 I	 have	 touched	upon	will	 not	 only	prevent	 our	wasting	 a	 lot	 of
time	in	trying	to	produce	impossible	forms,	but	will	lead	us	more	or	less	directly