# simple algebraic puzzle .....

Question asked by Shailaja Tiwari in #Brainy Puzzles on Sep 18, 2013

Shailaja Tiwari 路 Sep 18, 2013

Rank C1 - EXPERT

Q. xyz+xyz+xyz=zzz

Find the value of xyz?

condition: values of xyz must be same throughout the expression. Posted in: #Brainy Puzzles

Find the value of xyz?

condition: values of xyz must be same throughout the expression. Posted in: #Brainy Puzzles

Anoop Kumar 路 Sep 18, 2013

Rank A2 - PRO

x=1

y=8

z=5

y=8

z=5

Abhishek Rawal 路 Sep 18, 2013

Rank A2 - PRO

But that doesn't satisfies the equation.ianoopx=1

y=8

z=5

As per your answer, xyz=40, Hence LHS = 40+40+40 = 120. RHS = zzz = 5^3 = 125.

LHS != RHS

Anoop Kumar 路 Sep 18, 2013

Rank A2 - PRO

185+185+185 = 555 馃槢

Abhishek Rawal 路 Sep 18, 2013

Rank A2 - PRO

Bloody hell!

I thought question is : x*y*z + x*y*z + x*y*z = z*z*z [^_^]

I thought question is : x*y*z + x*y*z + x*y*z = z*z*z [^_^]

rahul69 路 Sep 19, 2013

Rank B2 - LEADER

If that is the case, then :Abhishek RawalBloody hell!

I thought question is : x*y*z + x*y*z + x*y*z = z*z*z [^_^]

x=3;

y=4;

z=6;

xyz+xyz+xyz=(3*4*6+3*4*6+3*4*6)=216=6*6*6=z*z*z

馃槒

Shailaja Tiwari 路 Sep 19, 2013

Rank C1 - EXPERT

Wow ,great you're absolutely right @ianoop.But how did you get this answer .I mean the logic if you could share would be of great help .I'd tried solving the question in several ways(including trial and error method too but couldn't get it .)

Such questions are generally asked in BARC's RMO.

Such questions are generally asked in BARC's RMO.

Abhishek Rawal 路 Sep 19, 2013

Rank A2 - PRO

@Shailaja Tiwari

3xyz = zzz => xyz = zzz/3

Take value of zzz = 111,222,333 upto 999.

Only one will satisfy the condition & that is 555.

Hence 555/3 = 185

So, x=1, y=8 & z=5

3xyz = zzz => xyz = zzz/3

Take value of zzz = 111,222,333 upto 999.

Only one will satisfy the condition & that is 555.

Hence 555/3 = 185

So, x=1, y=8 & z=5

Anoop Kumar 路 Sep 19, 2013

Rank A2 - PRO

I started from 3 digit number multiplied by 3 = zzz

start from unit place.

3*z should have z at the unit place.

start from possibilities 3*1=3, 3*6=18 you can come up with 3*5 = 15.

now you got 5 at unit place and carry 1 that means now you need a number with

3*y should have 4 at unit place, so adding a carry 1 will result 5.

and same way you can find value of x.

I hope this way is simple.

May be there Could be other simpler approach.

start from unit place.

3*z should have z at the unit place.

start from possibilities 3*1=3, 3*6=18 you can come up with 3*5 = 15.

now you got 5 at unit place and carry 1 that means now you need a number with

3*y should have 4 at unit place, so adding a carry 1 will result 5.

and same way you can find value of x.

I hope this way is simple.

May be there Could be other simpler approach.

boombastik85 路 Oct 21, 2015

Rank E2 - BEGINNER

Rahul 69 is right! xyz is x*y*z, 185 is an interesting answer but does not fit into rules of algebra