Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
my solution is shorter or not, it's for you to judge , but you may try this :
+let abc( 1<=a,b,c<=9) be the representative of such numbers. abc is odd, thus we have 300 odd numbers from 700-999
+when a=b=c , we have abc= 111c . There're 2 such numbers in all (777,999)
+when a=b<> c , we have 700<=abc=110b+c<=999==> b can only be 7,8,9. For each case, c can be 1,3,5,7,9 since abc is odd. Thus there're 3*5=15 cases of c overall. Except for 777 and 999, we have 13 abc left
+when a<>b=c , we have 700<=abc= 100a+11c<=999====> a can only be 7,8,9.For each case, c can be 1,3,5,7,9. Thus we have 15 cases of c in all. Except for 777 and 999, we have 13 abc left.
+when a=c<>b, we have 700<=abc= 101c +b <=999 ====> c can only be 7,9 . For each case, b can be 1--->9. Thus we have 18 cases of abc. Except for 777 and 999, we have 16 left.
+when b=0, we have 15 a0c.
Overall, the number of abc is (150-2-13-13-16-15)= 91.
for Y we have (remember different numbers, so we can use 7,8,9) 6 numbers..
for Z we have to have an odd digit, so 5 of them
so 3*6*5 +1 (since its says inclusive)=91.....
thanks for this explanation! i totally didn't consider using a combination. i'm having a little trouble understanding how you got 6 numbers for Y if you can use 7,8, or 9. do you mean cannot? if you can use 7,8, and 9 isn't the number of choices 9? (1,2,3,4,5,6,7,8,9) am i missing something?
for Y we have (remember different numbers, so we can use 7,8,9) 6 numbers..
for Z we have to have an odd digit, so 5 of them
so 3*6*5 +1 (since its says inclusive)=91.....
thanks for this explanation! i totally didn't consider using a combination. i'm having a little trouble understanding how you got 6 numbers for Y if you can use 7,8, or 9. do you mean cannot? if you can use 7,8, and 9 isn't the number of choices 9? (1,2,3,4,5,6,7,8,9) am i missing something?
You cannot use 7,8 and 9 for the X digit... since all digits must be distinct..
so 7yz up to 999...thus, you can only use numbers below 7, otherwise the three digits would not be distinct. 76Z would be fine
still don't get it [#permalink]
05 Oct 2005, 09:07
I'm doing something wrong cause I get an answer htat isn't there- 102
Take the possibilities for the 700s:
-For 70 there are 4 possibilites for the one's place (1,3,5,9)
-For the 71 there are 3 (3,5,9)
-for 72 there are 4 again (1,3,5,) and so on
-when you get to the 77 option- don't consider any of these possible
For the 800s:
80: 5 possibilities (1,3,5,7,9)
81: 4 possibilities (3,5,7,9)
and so on- exclude all the 88 options
for the 900's
same exact number as for the 700s
So for different possibilities considering the different number options in the 10's place:
0: 13 total
1: 10 total
2: 13 total
3: 10 total
4: 13 total
5: 10 total
6: 13 total
7: 7 total
8: 8 total
9: 7 total
That adds up to 102 right? I've got to be missing something silly but can't see it.
First I do not understand why to add one because the stem says that 700 and 699 are included since neither of those numbers are odd with no numbers repeated.
Second, we cannot exclude 7,8 and 9. for instance, 789 is a valid number.
Third: we do not have five choices for the last number, if the first number is 7, we cannot choose 7 for the last one.
If we take numbers from 700 to 799 and use this reasoning:
1*9*5 +1 =46 what is wrong.
In summary, I think it is a mere coincidence that we get the right result from that reasoning.