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3 digit integer pnc

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by advita » Thu Jan 20, 2011 6:18 am
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.
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by Rahul@gurome » Thu Jan 20, 2011 6:38 am
Refer to this post: https://www.beatthegmat.com/incorrect-og ... tml#325918

The problems are almost same.
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by stormier » Thu Jan 20, 2011 9:50 am
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.
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