Is there more than one way to multiply #25 xx 14# ?

2 Answers
Dec 16, 2017

The product is #350#, but there are several ways to get that answer.

Explanation:

Vertical Multiplication
#{:(" ",color(white)(x)25),(xx,underline(color(white)(x)14)),(" ",100),(" ",underline(250)),(" ",350):}#

That is the same as

#25 xx 14 = 25 xx (4+10) = (25xx4)+(25xx10) = 100+250 = 350#

We could instead use

#25 xx 14 = (20+5)xx14 = (20xx14)+(5xx14)= 280+70=350#
(Although the two multiplications will be challenging for many.)

Or we could use Factoring and re-arrange the factors

#25xx14 = 25xx(2xx7) = (25xx2)xx7 = 50 xx 7#

# = (5xx10)xx7 = (5xx7)xx10 = 35xx10 = 350#

Or

#25xx14 =(5xx5)xx(2xx7) = (5xx2)xx(5xx7) = 10xx35=350#

There are other ways of factoring, but I think those two are the most helpful.

Dec 16, 2017

Yes

Explanation:

Here are just a few...

Long multiplication

#25color(white)(00000)larr 25 xx 10#
#underline(100)color(white)(0000)larr 25 xx 4#
#350#

or:

#28color(white)(00000)larr 14 xx 20#
#underline(color(white)(0)70)color(white)(0000)larr 14 xx 5#
#350#

Russian multiplication

#25color(white)(000)color(red)(14)#
#color(grey)(12)color(white)(000)color(grey)(28)#
#color(white)(0)color(grey)(6)color(white)(000)color(grey)(56)#
#color(white)(0)3color(white)(00)color(red)(112)#
#color(white)(0)1color(white)(00)underline(color(red)(224))#
#color(white)(0000)350#

The left hand column is formed by repeatedly halving and discarding any remainder. The right hand column is formed by repeatedly doubling. Then ignoring any row where the number in the left hand column is even, add up the numbers in the right hand column.

Squares and difference

#25xx14 = ((25+14)/2)^2 - ((25-14)/2)^2#

#color(white)(25xx14) = 1521/4-121/4#

#color(white)(25xx14) = 1400/4#

#color(white)(25xx14) = 350#

(Note this method works better when both multiplicands are odd or both even)