Can you solve these math problems?

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

  1. aaron7e7 macrumors newbie

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    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

    Chaszmyr

    Joined:
    Aug 9, 2002
    #2
    To answer your question directly, B represents 3.

    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

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    #3
    nice!! what about this one...
    The last decimal digit of 2004^2004 is..
    A-0 B-2 C-4 D-6 E-8
     
  4. jsw Moderator emeritus

    jsw

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    Andover, MA
    #4
    Sorry - but 4 doesn't work. :(
     
  5. HexMonkey Administrator

    HexMonkey

    Staff Member

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    #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

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    #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

    steelphantom

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    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. :eek:
     
  8. Abstract macrumors Penryn

    Abstract

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    #8
    D-6
     
  9. HexMonkey Administrator

    HexMonkey

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    #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.
     
  10. Chaszmyr macrumors 601

    Chaszmyr

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    #10
    My bad. Guess I didn't read carefully enough.
     
  11. Abstract macrumors Penryn

    Abstract

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    #11
    My advice to everyone is to never argue about maths with a guy named HexMonkey.
     
  12. Don't panic macrumors 603

    Don't panic

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    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
     

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