Maths Olympiad Challenge!!!

Discussions on tuition centres/enrichment services that specialise in Mathematics.

Maths Olympiad Challenge!!!

Postby Maths Hub » Fri Oct 22, 2010 1:03 am

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 :D !

Maths Hub
BlueBelt
BlueBelt
 
Posts: 368
Joined: Wed Oct 20, 2010 11:52 pm
Total Likes: 0


Re: Maths Olympiad Challenge!!!

Postby Jojo_2010 » Fri Oct 22, 2010 1:35 pm

7 X ABCDEF = 6 X DEFABC

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

7X6X5X4X7X2X3=6X7X2X3X6X5X4=30240.

Am i rite?

Jojo_2010
GreenBelt
GreenBelt
 
Posts: 122
Joined: Wed Aug 04, 2010 11:48 pm
Total Likes: 0


Re: Maths Olympiad Challenge!!!

Postby Maths Hub » Fri Oct 22, 2010 5:53 pm

Jojo_2010 wrote: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.
Last edited by Maths Hub on Fri Oct 22, 2010 9:40 pm, edited 1 time in total.

Maths Hub
BlueBelt
BlueBelt
 
Posts: 368
Joined: Wed Oct 20, 2010 11:52 pm
Total Likes: 0


Re: Maths Olympiad Challenge!!!

Postby mujin » Fri Oct 22, 2010 8:53 pm

Maths Hub wrote: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 :D !


Let me try:

ABCDEF = 461538 :D

mujin
YellowBelt
YellowBelt
 
Posts: 12
Joined: Thu Jul 29, 2010 12:03 am
Total Likes: 0


Re: Maths Olympiad Challenge!!!

Postby mujin » Fri Oct 22, 2010 9:00 pm

mujin wrote:
Maths Hub wrote: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 :D !


Let me try:

ABCDEF = 461538 :D



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 :D :D

mujin
YellowBelt
YellowBelt
 
Posts: 12
Joined: Thu Jul 29, 2010 12:03 am
Total Likes: 0



Re: Maths Olympiad Challenge!!!

Postby iFruit » Fri Oct 22, 2010 10:03 pm

mujin wrote:
mujin wrote:
Maths Hub wrote: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 :D !


Let me try:

ABCDEF = 461538 :D



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 :D :D


Nice One :)

iFruit
BlueBelt
BlueBelt
 
Posts: 217
Joined: Thu Sep 09, 2010 11:20 am
Total Likes: 0


Re: Maths Olympiad Challenge!!!

Postby Maths Hub » Fri Oct 22, 2010 11:27 pm

iFruit wrote:
mujin wrote:
mujin wrote:
Maths Hub wrote: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 :D !


Let me try:

ABCDEF = 461538 :D



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 :D :D


Nice One :)


Well done! This is indeed correct!

Maths Hub
BlueBelt
BlueBelt
 
Posts: 368
Joined: Wed Oct 20, 2010 11:52 pm
Total Likes: 0


Postby Guest » Fri Oct 22, 2010 11:44 pm

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... :wink:

6994 ABC = 5993 DEF

538 ABC = 461 DEF

:sweat: :sweat: :sweat:
Guest
 

Postby Maths Hub » Fri Oct 22, 2010 11:47 pm

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

Maths Hub
BlueBelt
BlueBelt
 
Posts: 368
Joined: Wed Oct 20, 2010 11:52 pm
Total Likes: 0


Postby ChiefKiasu » Sat Oct 23, 2010 12:08 am

Maths Hub wrote:Pls refer to http://www.kiasuparents.com/kiasu/forum ... hp?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?

ChiefKiasu
Site Admin
 
Posts: 15221
Joined: Mon Sep 03, 2007 9:16 am
Location: Singapore
Total Likes: 319


Next

Return to Mathematics