Maths Olympiad Challenge!!!

[Maths Hub]

Hi Parents and Students,


Here's a Challenging question for all of you to solve:

Find the 6 digit number ABCDEF
such that 7 x ABCDEF = 6 x DEFABC where DEFABC is another 6 digit number.

Different letters stand for different single-digit whole numbers.

Have a fun time solving !

[Jojo_2010]

7 X ABCDEF = 6 X DEFABC


A=6
B=5
C=4
D=7
E=2
F=3

7X6X5X4X7X2X3=6X7X2X3X6X5X4=30240.

Am i rite?

[Maths Hub]

Jojo_2010:

7 X ABCDEF = 6 X DEFABC


A=6
B=5
C=4
D=7
E=2
F=3

7X6X5X4X7X2X3=6X7X2X3X6X5X4=30240.
Am i rite?

Not Correct
For your equation above, LHS is not equal to RHS.

Anyway, ABCDEF is a 6 digit number, not a product of 6 single digit number.

[mujin]

Maths Hub:

Hi Parents and Students,


Here's a Challenging question for all of you to solve:

Find the 6 digit number ABCDEF
such that 7 x ABCDEF = 6 x DEFABC where DEFABC is another 6 digit number.

Different letters stand for different single-digit whole numbers.

Have a fun time solving !

Let me try:

ABCDEF = 461538

[mujin]

mujin:

Maths Hub:

Hi Parents and Students,


Here's a Challenging question for all of you to solve:

Find the 6 digit number ABCDEF
such that 7 x ABCDEF = 6 x DEFABC where DEFABC is another 6 digit number.

Different letters stand for different single-digit whole numbers.

Have a fun time solving !

Let me try:

ABCDEF = 461538

My working :

ABCDEF x 7 = DEFABC x 6

[(ABC x 1000) + DEF] x 7 = [(DEF x 1000) + ABC] x 6

7000 ABC + 7 DEF = 6000 DEF + 6 ABC

6994 ABC = 5993 DEF

538 ABC = 461 DEF

538 (461) = 461 (538)

whereby (461) = (ABC) and
(538) = DEF

so.. ABCDEF = 461538

Hope I am right

[iFruit]

mujin:

mujin:

[quote=\"Maths Hub\"]Hi Parents and Students,


Here's a Challenging question for all of you to solve:

Find the 6 digit number ABCDEF
such that 7 x ABCDEF = 6 x DEFABC where DEFABC is another 6 digit number.

Different letters stand for different single-digit whole numbers.

Have a fun time solving !

Let me try:

ABCDEF = 461538

My working :

ABCDEF x 7 = DEFABC x 6

[(ABC x 1000) + DEF] x 7 = [(DEF x 1000) + ABC] x 6

7000 ABC + 7 DEF = 6000 DEF + 6 ABC

6994 ABC = 5993 DEF

538 ABC = 461 DEF

538 (461) = 461 (538)

whereby (461) = (ABC) and
(538) = DEF

so.. ABCDEF = 461538

Hope I am right

:D[/quote]Nice One

[Maths Hub]

iFruit:

mujin:

[quote=\"mujin\"]
Let me try:

ABCDEF = 461538

My working :

ABCDEF x 7 = DEFABC x 6

[(ABC x 1000) + DEF] x 7 = [(DEF x 1000) + ABC] x 6

7000 ABC + 7 DEF = 6000 DEF + 6 ABC

6994 ABC = 5993 DEF

538 ABC = 461 DEF

538 (461) = 461 (538)

whereby (461) = (ABC) and
(538) = DEF

so.. ABCDEF = 461538

Hope I am right

Nice One :)[/quote]Well done! This is indeed correct!

[corneyAmber]

Not easy to see that both sides divisible by 13 to reduce to 538 and 461 leh.....at least not for old mummy like me...


6994 ABC = 5993 DEF

538 ABC = 461 DEF

[Maths Hub]

Pls refer to http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=14953 for the next Maths Olympiad Challenge!!!

[ChiefKiasu]

Maths Hub:

Pls refer to http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=14953 for the next Maths Olympiad Challenge!!!

Umm... seems like it is not much of a challenge for the super bright parents and kids we have here right on KiasuParents.com. Maybe need to turn up the difficult by a couple of dozen notches?