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Ok so this is a gmatclub test but I don't get the explanation. Can someone help?

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

40 60 72 81 104

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

I would post another way with little short method. First of all let's find out all numbers greater then 800 and having all different digits. For hundreth place you have 2 choices, for tenth digit you have (10-1)=9 choices as one digit out of 10 have already used in hundreth place, for unit's digit you have (10-2) = 8 choices Total three digit numbers with all different digits greater then 800 = 2*8*9 = 144 As 144 is an even number, half of these will be odd and half of this will be even. So answer will be 72 Ans: C (72)

Alright so you need odd number so the first odd number is 801 and the last odd number is that is different is 987

there a total of {(987-801)/2} + 1 odd numbers. There are 94 odd numbers. Now you need to get rid of odd numbers that violate the rule that each digit must be different. there a total of 5 odd numbers digits per group of tens - >1,3,5,7,9. From 801 to 879 you count all the odds till this point but you discard all the odd in the 80s group which there are 5 -> 81, 83,85,87,89 - and you discard one odd in 50s -> 855 . The 900 forces you to make an additional consideration. There is one less odd per group of ten since you'll have a repeat of 9 - example 909, 919. There are only 4 odd numbers per group of ten and you also get rid of 955. There are a total 8 numbers that you exclude in the 900s on the way up to 987.

So the total number of odds you discard is 13

So the answer is 94 -13 = 81

IF that helped a kudo would be nice
_________________

Ok so this is a gmatclub test but I don't get the explanation. Can someone help?

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

40 60 72 81 104

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.

wow amazing approach ... now this is crystal clear !!
_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

So, we have 1[x]*5[z]*(10-2)[y] = 40 integers that begin with 8 and 1[x]*4[z]*(10-2)[y] = 32 integers that begin with 9. N = 40 + 32 = 72
_________________

Hey Bunuel, in the solution below, I did not understand the highlighted part.

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900: 1 choice for the first digit: 8; [highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight] 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; [highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight] 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Hey Bunuel, in the solution below, I did not understand the highlighted part.

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900: 1 choice for the first digit: 8; [highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight] 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; [highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight] 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.

Odd numbers should end with odd digit.

Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.

Ok so this is a gmatclub test but I don't get the explanation. Can someone help?

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

40 60 72 81 104

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.

Bunuel, Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.

Please do share your thoughts on this.

This was my approach to solve this problem

x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).

Of course multiplying these choices does not lead to any of the answer choice.
_________________

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Bunuel, Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.

Please do share your thoughts on this.

This was my approach to solve this problem

x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).

Of course multiplying these choices does not lead to any of the answer choice.

There are 5 choices for units digit when hundreds digit is 8 and 4 when hundreds digit is 9. So you should count numbers in these 2 ranges separately.
_________________

Hey Bunuel, in the solution below, I did not understand

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. --> this should be 9 digits right why havent u considered 0

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. -->-> this should be 9 digits right why havent u considered 0

Hey Bunuel, in the solution below, I did not understand

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. --> this should be 9 digits right why havent u considered 0

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. -->-> this should be 9 digits right why havent u considered 0

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.

Odd numbers should end with odd digit.

Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.

Ok so this is a gmatclub test but I don't get the explanation. Can someone help?

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

40 60 72 81 104

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. ---> I am confused here, how did we get 8? is it 10 - 1 - 1 ? What does first digit and third digit means ?

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.

I am confused with the selection of second digit.

3 digit integer abc: we used 1 digit for the first digit \(a\), 1 digit for the third digit \(c\), so for the second digit \(b\) there are 10-1-1=8 digits left.
_________________

I did not understand why you included 900 again in the range of 900 - 999 (should it not be 901 to 999). Why we counted 900 two times?

[highlight]In the range 800 - 900:[/highlight]

1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

I did not understand why you included 900 again in the range of 900 - 999 (should it not be 901 to 999). Why we counted 900 two times?

In the range 800 - 900:

1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999:

1 choice for the first digit: 9;

As we are counting ODD numbers it doesn't make any difference whether we include even number 900 once, twice or even ten times.
_________________

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