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@Shailaja Tiwari • 18 Sep, 2013 • 1 like

Q. xyz+xyz+xyz=zzz

Find the value of xyz?

condition: values of xyz must be same throughout the expression.

Find the value of xyz?

condition: values of xyz must be same throughout the expression.

@Anoop Kumar • 18 Sep, 2013
x=1

y=8

z=5

y=8

z=5

@Abhishek Rawal • 18 Sep, 2013

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

LHS != RHS

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 • 18 Sep, 2013
185+185+185 = 555 😛

@Abhishek Rawal • 18 Sep, 2013
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 • 19 Sep, 2013 • 3 likes

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

😏

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

😏

@Shailaja Tiwari • 19 Sep, 2013
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 • 19 Sep, 2013 • 5 likes
@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 • 19 Sep, 2013 • 1 like
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 • 21 Oct, 2015
Rahul 69 is right! xyz is x*y*z, 185 is an interesting answer but does not fit into rules of algebra

@Divisha Madupalli • 12 Feb, 2020

3z = z or 1z or 2z or 3z and so on

But only z and 1z with satisfy the equation as z has to be between 0 and 9.

=> 3z = z OR 3z = 10 + z

=> z = 0 OR z = 5

=> z=5

Now,

xyz + xyz + xyz = 555

=> xyz = 185

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