牛人求救看看这道奇怪的数学题-OG 解释 不清晰

liuyongmei

11. Of the three-digit integers greater than 700, how many
have two digits that are equal to each other and the
remaining digit different from the other two?
(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

In three-digit integers, there are three pairs of
digits that can be the same while the other digit is
diff erent: tens and ones, hundreds and tens, and
hundreds and ones. In each of these pairs, there
are 9 options for having the third digit be
diff erent from the other two. Th e single exception
to this is in the 700–799 set, where the number
700 cannot be included because the problem
calls for integers “greater than 700.” So, in the
700–799 set, there are only 8 options for when
the tens and ones are the same. Th is is shown in
the table below.

Number of digits available for the third digit
when two given digits are the same
Same      701–799 800–899 900–999
tens and ones  8      9       9
hundreds and tens 9   9      9
hundreds and ones 9   9      9


Th us, of the three-digit integers greater than 700,
there are 9(9) – 1 = 80 numbers that have two
digits that are equal to each other when the
remaining digit is diff erent from these two.
The correct answer is C.


我看不明白： 如果已经按照701–799这样的分法了，如何能得出tens and ones similar?毕竟这一组数里面百位数就是7。如果十位数，个位数再相同，那么就成了3位数字都一样了。

大家

liuyongmei

谢谢大家，我发现我理解错题意了。以为是一对数，相互有2个digit相同 。

yojo88111

这道题做过。。。同意楼上的~
就是700 除了77+  里面排除777   其他的都是一样啊 012345689 算呗！

kevinkelvin

当百位数字为7，且十位与个位equal的时候，十位与个位不能同时取7，又因为这个数字要greater than 700，所以不能同时取0，那么只能同为1 2 3 4 5 6 8 9，一共8个数字。