HMMT 十一月 2011 · GEN 赛 · 第 8 题

英文解析

8. [ 6 ] Find the number of integers x such that the following three conditions all hold:
   • x is a multiple of 5
   • 121 < x < 1331
   • When x is written as an integer in base 11 with no leading 0 s (i.e. no 0 s at the very left), itsrightmost digit is strictly greater than its leftmost digit.
   Answer: 99 We will work in base 11, so let x = def such that d > 0. Then, based on the firsttwo conditions, we aim to find multiples of 5 between 100 and 1000 . We note that 11
   11 11
   def ≡ 11 · d + 11 · e + f ≡ d + e + f (mod 5) .2
   Hence, x a multiple of 5 if and only if the sum of its digits is a multiple of 5. Thus, we wish to find 11
   triples ( d, e, f ) with elements in 0 , 1 , 2 , · · · , 9 , 10 such that d + e + f ≡ 0 (mod 5) and 0 < d < f .
   Note that if we choose d and f such that d < f , there is exactly one value of e modulo 5 that wouldmake d + e + f ≡ 0 (mod 5). Once the this value of e is fixed, then there are two possibilities for eunless e ≡ 0 (mod 5), in which case there are three possibilities. Thus, our answer is twice the numberof ways to choose d and f such that 0 < d < f plus the number of ways to choose d and f such that
   General Testd + f ≡ 0 (mod 5) and 0 < d < f (to account for the extra choice for the value of e ). Note that the
   ( )
   number of ways to choose 0 < d < f is just since any any choice of two digits yields exactly one 10
   way to order them. The number of ways to choose d + f ≡ 0 (mod 5) and 0 < d < f can be found by 2
   listing: ( d, f ) = (1 , 4) , (1 , 9) , (2 , 3) , (2 , 8) , (3 , 7) , (4 , 6) , (5 , 10) , (6 , 9) , (7 , 8), for 9 such pairings.
   ( )
   Hence, the total is 2 + 9 = 99 possibilities for x .10 2