1. Ok, lets see... first digit pretty much has to be 2, by my count, else you risk going over.

4 times what equals *2... 8 off the top of my head with 32, so last digit must be 8.

A and E are most likely done then.

so 23758... 32, carry the 3.
20 + 3 = 23, gives us a 3, carry the 2... so far so good.
7 time 4 plus 2 is 30, which is a 0. Doesn't work. Middle needs to equal itself in the end. Hmm... used 3, 4 no, used 5, 6 works. So 6 times 4 plus 2 is 26... good.
4 times 3, plus 2 is 14... shit.

ok, that didn't work.
*thinks*
[15 minutes later]
ok, got it. Same steps...
21978.
32, 2 carry 3.
28 and 3 is 31, 1 carry 3.
36 and 3 is 39, 9 carry 3.
4 and 3 is 7, 7 no carry.
Finally 8.

A=2
B=1
C=9
D=7
E=8
Confirm?

2. Senior Member Community Builder
Join Date
Oct 2005
Location
FL
Posts
2,785
Can't expect no less from you Volvogga!

But you took the long way around, there is a easier way.
I'll post just for the sake of posting:

Its obvious that A can't be more then 2, A also has to be an even number, because its a product of a even number(4). So a has to be 2.

Now E, E*4 has to end in 2, only number that satisfy that are 3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number that satisifies both conditions

Next B, we know that 2BCD8*4 is at least 80000 and less than 90000. B can not be more than 2 otherwise 2BCD8 * 4 would be more than 80000. we used 2 already so B must be 0 or 1. Consideing B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies this condition. So B must be 1.

D8*4 must end in the digits 12. The only possiblity is D=7 (78*4=312).

Lastly for C:
21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300 = 9
@adonai: You said "It can't be 2 since E must then be 8 or 9, 8*4=32, 9*4=36, neither of which ends in 2."
But 8*4 = 32 < ends in 2 XD

3. Senior Member Community Builder
Join Date
Aug 2006
Posts
5,630
Originally Posted by irecinius
@adonai: You said "It can't be 2 since E must then be 8 or 9, 8*4=32, 9*4=36, neither of which ends in 2."
But 8*4 = 32 < ends in 2 XD
Oh, right, don't know what I was thinking there.

4. Ah hell... that'd a been abit easier.

Oh well, how about some Euler (hint)?

What's φ(9) (that's phi(9))?

5. Senior Member Always Around
Join Date
Jul 2007
Location
somewhere but not here
Posts
1,968
hahaha, I just learn this in one of my math classes

φ(9) = φ(3^2) = 9 * (1 - 1/3) = 6

6. Bingo. Could also use relative primes and GCD. ^^

7. Senior Member Always Around
Join Date
Jul 2007
Location
somewhere but not here
Posts
1,968
... sorry for not posting a problem earlier, i am in exam week :S
now here is another one:

John sells 1000 and he has left more than half of what he originally had. If he then sells 502 he would have left less than 500. How many books did he originally had?

8. Senior Member Community Builder
Join Date
Oct 2005
Location
FL
Posts
2,785
(x - 1000) - 502 < 500
-1502+x < 500
x < 2002

e Se
x-1000 > 1/2x
x > 1/2x + 1000
1/2x > 1000
x > 2000
Since x > 2000 and x < 2002, so: John had 2001 books...

Proving:
2001 - 1000 = 1001, that is indeed more then 2001/2 (that is 1000.50)
1001 - 502 = 499, that is indeed less then 500.

It was that easy? or I missed the point completely?

9. Senior Member Always Around
Join Date
Jul 2007
Location
somewhere but not here
Posts
1,968
yeap, you got it right

10. Senior Member Community Builder
Join Date
Oct 2005
Location
FL
Posts
2,785
Humm...
Question then!

How many are we?
Knowing that the probability that at least two of us have birthdays on the same day is less than half, but that this would not be the case were we one more in number.

Been 2 days... I guess people just didn't had time to sit down a bit and work on the problem.. I'll give a BIG hint.

364/365
= Probability
Last edited by irecinius; 02-17-2008 at 12:51 AM. Reason: Tip: don't account leap year...

Page 4 of 7 First ... 23456 ... Last

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
vBulletin Skin by: ForumThemes.com