IIT JEE , AIEEE Forum - Mathematics , Algebra - CHALLENGE FOR ALL -TRICK

GoNik

CHALLENGE FOR ALL -TRICK
SUM TRICK, A MAGIC
Friends during a recent visit to a magic show, a person was asked to think of a

3-digit number, whose 100th digit was bigger than the rest numbers, for ex. 971 or

956 i.e. 9 > 5,6. And all the numbers should be distinct. Then the algorithm of the

magic went like this :
1- Take the number and from this number subtract the number obtained by

reversing the digits,
2- Now having this number add this to the number obtained by reversing its digits

(the new one, the above one in 2-part),
3- The number obtained will always be 1089!!! and i was shocked how could it be?

Funny isn't it but my question is that can anyone prove this with a mathematical

base. I tried it with a good success., its a challenge to all.
This example will clear any doubt in the algorithm:
Let a number be 871, the reverse number is 178. And now, 871-178=693.
Now reverse of 693 is 396. So 693+396=1089
It is believed that this trick was derived from a Russian Olympiad where students

were asked to prove this but only a handful of them could prove it.

indr12365

gr 2 see some one putting genuine challenges like tat..thumbs up!!!

GoNik

tanx but wats the reply gal? hmmmmmmm

RyuAmakusa

I GOT IT BUT IT TAKES TIME TO POST. SO PLZ. WAIT. AND IT INVOLVES JUST VERY BASIC IDEA OF ADDITION AND SUBRACTION.

computer001

let the number be abc ie a*100 + b*10 +c
now reverse is c*100 + b*10 + a
diff is 100(a-c) + c-a= 100(a-c-1) + 100 + c-a(note c-a -s negative)
so v have 100(a-c-1) + 90 + (10 + c-a)
the last term in this is obviously <10...
now this is of the form xyz=x*100 + y*10 + z where x= (a-c-1)
y=9 z=(10 + c-a)
now reverse of num is z*100 + y*10 + x
the sum ofthe prev 2 numbers is (x+z)*100 +2*9*10+ z+x= 900+180+9=1089

RyuAmakusa

let the no be (100)x + (10)y + z now x>y>z.

the reversed no. is 100z + 10y + x now subract.

100x + 10y + z

(-) 100z + 10y + x

now if u do normal sub. z-x will be - so what do we do recall from the 1 st standard. borrow.

100x + 10(y-1) +10+ z

(-) 100z + 10y + x even now u need to borrow.

100(x-1) +100+ 10(y-1) +10+ z

(-) 100z + 10y + x now sub.

= (100( x-1-z)) + (100 -10 )+ (10 +z -x)

// i have enclosed each digit in brackets.

= (100( x-1-z)) + (10(9) )+ (10 +z -x) now reverse this no.

(100(10 +z -x)) + (10(9) ) + ( x-1-z) now add.

= (100( x-1-z +10 +z -x) +90 + 90 +(10 +z -x + x-1-z )

= 100 *9 +180 + 9

= 1089!!!

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