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  1. Prime numbers, perhaps? Then all four must be odd, since no even number greater than 2 is prime. And none of them can be 5, because any number ending in 5 has 5 as a factor. So they are 1, 3, 7, 9.

    2 Answers · Science & Mathematics · 28/10/2007

  2. ... AAAD AABA AABB AABC AABD AACA AACB AACC AACD AADA AADB AADC...

    3 Answers · Science & Mathematics · 09/10/2010

  3. ...or 9. Utilising these facts, the only possibilities for AACA are : 1131, 1171, 1191, 7717, 7737 and 7797. Of these, only 1171 and 7717...

    5 Answers · Science & Mathematics · 01/11/2007

  4. ... start trial with digit 1 = A, In the first number AACA, C has to be 7 as 3 or 9 will make it divisible by 3. In the...

    1 Answers · Science & Mathematics · 23/10/2007

  5. ...similar big expansion, only much longer: aaaa+aaab+aaac+aaad+aaba+aabb+aabc+aabd+aaca+aacb+... If you think about it for a while, you could certainly list all the ...

    2 Answers · Science & Mathematics · 22/08/2008

  6. ... AAAB AAAC AABA AABB AABC AACA AACB AACC ....etc. Formula gives: cover(n,k...

    3 Answers · Science & Mathematics · 03/07/2008

  7. a b c or d cannot be equal to a number that is divisible by two or 5 so they can on;y be 1,3,7,9 therefore by taking this into account and substituting different values the only combination that satisfies this criteria is a=1 b=3 c=7 d=9...

    1 Answers · Science & Mathematics · 23/10/2007

  8. When you have n objects to choose from and you have k slots to fill, you can do it in n^k ways. For n=k=4, that is 4^4 = 256 possible combinations. For n=k=5, that is 5^5 = 3125 possible combinations.

    4 Answers · Science & Mathematics · 11/07/2016

  9. ...eaaa eaab eaac eaba eabe eaca caaa caab caac caba cabe caca aaaa aaab aaac aaba aabe aaca abaa abab abac abea acaa acab acac baaa baab baac baba babe baca beaa beab beac

    2 Answers · Science & Mathematics · 09/03/2009

  10. You don't need to ignore the wild cards, and you don't calculate it the way you were doing. Suppose there are 14 cards of each type (soldier, horse, cannon) and 2 wild cards. The only way to need five cards is to be unable to trade...

    5 Answers · Science & Mathematics · 17/11/2007

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