# Can you solve these math problems?

Discussion in 'Community Discussion' started by aaron7e7, Mar 3, 2006.

1. ### aaron7e7 macrumors newbie

Joined:
Jan 1, 2005
Location:
Huntington, Ut
#1
In my high school we had the option of going to a state math competition. But we had to take a qualifier. I didn't feel like taking this test but my math teacher showed us this test after it was over and explained some of the problems to us. I just wanted to see if some of you could figure these out. Here is one of them:

1. The word WEBER hides a 5-digit number. Different letters indicate different digits, and same letters stand for the same digits. Every digit is a prime number and so is the sum of the 5 digits. The 2-digit number EW and the 3-digit number EBR are also primes. What digit does letter B represent.

(A) 1 (B) 2 (C) 3 (D) 5 (E) 7

If you want some more problems then just tell me and i will post some.

2. ### Chaszmyr macrumors 601

Joined:
Aug 9, 2002
#2

W=7
E=4
B=3
E=4
R=1

W+E+B+E+R=19

EBR = 431

EW = 47

I hate math puzzles, they aren't fun but I feel inclined to figure them out

3. ### aaron7e7 thread starter macrumors newbie

Joined:
Jan 1, 2005
Location:
Huntington, Ut
#3
The last decimal digit of 2004^2004 is..
A-0 B-2 C-4 D-6 E-8

4. ### jsw Moderator emeritus

Joined:
Mar 16, 2004
Location:
Andover, MA
#4
Sorry - but 4 doesn't work.

### Staff Member

Joined:
Feb 5, 2004
Location:
New Zealand
#5
There are 4 letters representing 4 digits (2, 3, 5 and 7 - the prime number digits). The sum must be odd in order to be prime, so 2 must appear twice in W+E+B+E+R (because even+even+odd+odd+odd=odd, but even+odd+odd+odd+odd=even), therefore E is 2. W and R must each be one of 3 and 7 so that EW and EBR are prime (2+ digit prime numbers can't end in 2 or 5). That leaves B as being 5.

2004^1 ends in 4
2004^2 ends in 6 (4*4 ends in 6)
2004^3 ends in 4 (6*4 ends in 4)
2004^4 ends in 6 (4*4 ends in 4)
Continuing the pattern, 2004^2004 ends in 6

6. ### aaron7e7 thread starter macrumors newbie

Joined:
Jan 1, 2005
Location:
Huntington, Ut
#6
i have no idea if you guys are right...i couldn't do any of these but i will get the answers on monday. Here is a another one.

Write down all the integers from 1 to 30 to form the number 1234567891011121314.....2930
Now, delete 44 digits from this number and call the resulting number N. What is the possible value of N that is closest (smaller, larger, or equal) to 5 x (times) 10^6

A-5,001,220 B-4,998,930 C-4,999,888 D-5,000,111 E-none of these

7. ### steelphantom macrumors 6502a

Joined:
Oct 15, 2005
#7
Not to mention that EW would equal 28, not 47. Now I'm sitting here trying to figure it out!

Edit: Now that I look at it, 47 is right in that case. I was thinking that EW meant E*W.

8. ### Abstract macrumors Penryn

Joined:
Dec 27, 2002
Location:
Location Location Location
#8
D-6

### Staff Member

Joined:
Feb 5, 2004
Location:
New Zealand
#9
Neither 4999 (from 4, 9, 19 and 29) or 5000 (from 5, 10, 20 and 30) can be formed while leaving 3 remaining digits after them, which rules out C and D. 4998 can be formed (from 4, 9, 19 and 28) leaving 2930 as the remaining digits. This leaves 4,998,930 (B) as the largest possible number below 5x10^6. Similarly, 5001 can be formed (from 5, 10, 20 and 21) leaving 22 to 30 remaining, which lets us form A (the smallest number that can be formed greater than 5x10^6). 4,998,930 is closer to 5x10^6 than 5,001,220, giving an answer of B.

Joined:
Aug 9, 2002
#10

11. ### Abstract macrumors Penryn

Joined:
Dec 27, 2002
Location:
Location Location Location
#11
My advice to everyone is to never argue about maths with a guy named HexMonkey.

12. ### Don't panic macrumors 603

Joined:
Jan 30, 2004
Location:
having a drink at Milliways
#12

interestingly enough, the answer (b=5) would be correct also if you adopted the wider (old) definition that includes 1 as a prime number (which would make hex's solution incorrect in method, because now you could have odd+odd+odd+odd+odd=odd).
With 1 'in', there are (at least?) 4 solutions (31517,37571,32521,32527) where in all cases b=5 (and the first two are prime number themselves, plus 37571 is a palindromic prime!)
Using the stricter (and correct) definition of prime the only (hex's) solution is 32527