Refer to this post: https://www.beatthegmat.com/incorrect-og ... tml#325918
The problems are almost same.
3 digit integer pnc
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advita wrote:Of the three-digit integers greater than 800, how many have two
digits that are equal to each other and the remaining digit
different from the other two?
answer:53
pl show your proceedings...thanks.
digits starting with 8 --> 88x + 8x8 + 8yy = 9 + 9 + 8 (y cannot be 0) = 26
digits starting with 9 --> 99x + 9x9 + 9yy = 9 + 9 + 9 =27
total = 26+27 =53.
In 8x8 form - x can be any number from 0 to 9, but not 8. Thus total of 9 choices. Similar argument for others.













