How many odd three-digit integers greater than 800 are there

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Wed May 02, 2018 2:32 pm

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How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

The OA is C.

Please, can anyone assist me with this PS question? Thanks in advance!

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

How many odd three-digit integers greater than 800 are ther

by regor60 » Thu May 03, 2018 5:29 am

BTGmoderatorLU wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

The OA is C.

Please, can anyone assist me with this PS question? Thanks in advance!

So the range is 801 to 999 inclusive. Start with the numbers in the 800's.

First digit is 8. Assume an even number as the second digit. Since 8 is taken there are 4 choices. Third digit has 5 choices. Total choices 1x4x5 = 20

Same thing but assume odd number as second digit. 5 choices. Third digit also odd, but now 4 choices since used one for the second digit. Total choices 1x5x4 = 20.

Now do the 900's in the same way. First digit 9. Second digit even, 5 choices. Third digit odd, 4 choices since 9 is taken, total choices 1x5x4 = 20.

Second digit odd. 4 choices since 9 taken. Third digit odd, 3 choices since two have been taken, total 1x4x3 = 12.

Adding the choices 20 + 20 + 20 + 12 = 72, C

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu May 03, 2018 5:48 am

BTGmoderatorLU wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

Case 1: Hundreds digit = 8
Number of options for the units digit = 5. (Any of the 5 odd digits.)
Number of options for the tens digit = 8. (Of the 10 digits, any but the two already used.)
To combine the options above, we multiply:
5*8 = 40.

Case 1: Hundreds digit = 9
Number of options for the units digit = 4. (Of the 5 odd digits, any but 9.)
Number of options for the tens digit = 8. (Of the 10 digits, any but the two already used.)
To combine the options above, we multiply:
4*8 = 32.

Total number of viable integers = Case 1 + Case 2 = 40 + 32 = 72.

The correct answer is C.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu May 03, 2018 6:23 am

Here's a similar one to practice with: https://www.beatthegmat.com/counting-a-3 ... 88550.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Fri May 04, 2018 9:32 am

BTGmoderatorLU wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 56
C. 72
D. 81
E. 104

Since the integer is odd, the last (or units) digit has five choices: 1, 3, 5, 7 or 9.

If the first (or hundreds) digit is 8, then the second (or tens) digit has 8 choices (since that digit has to be different from the hundreds and from the units digit). Thus, there are 1 x 8 x 5 = 40 odd integers in the 800s.

If the first (or hundreds) digit is 9, then the units digit has only 4 choices (since it can't be 9), and the second (or tens) digit has 8 choices (since that digit has to be different from the hundreds and from the units digit). Thus, there are 1 x 8 x 4 = 32 odd integers in the 900s.

Therefore, we have a total of 40 + 32 = 72 such integers.

Answer: C

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Head of GMAT Instruction
[email protected]

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