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Foreheadson
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How many odd three-digit integers greater than 800 are there such that 06 Jul 2021, 04:26
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Hi, I have used the following approach:

I have counted quantity of numbers with at least 2 similar digits.

1) in 800s for every next 10 numbers (except for 880s) we have 2 numbers with similar digits. for example 811 and 818. 800 and 808 also, but we are told greater than 800, so 800. so there are 810s, 820, 830 ... 870s, 890s, 900s, ... 980s. so 17*2=34. +1 from 808 = 35. Then add 10 from 880s, and 10 from 990s. 35+10+10=55.

2) there are total of 199 numbers. So 199-55=144. choose only odd numbers, 144/2=72.
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How many odd three-digit integers greater than 800 are there such that 14 Mar 2024, 21:11
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I get 45 options for the 800-899 range..
8(EVEN)X - X has 1,3,5,7,9 = 5 options. 5 EVEN * 5 Options = 25
8(ODD)X - X has 1,3,5,7,9 initially, but since they cant repeat, we will have 4 options for X for every ODD number in tenths place. 5 ODD * 4 Options = 20.

Total is 45.
Bunuel
How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?

A. 40
B. 60
C. 72
D. 81
E. 104


In the range 800 - 899:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)

8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.

Total for this range: \(1*5*8 = 40\).

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is not an option, as it's the first digit)

8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.

Total for this range: \(1*4*8 = 32\).

Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).


Answer: C

Hope it's clear.
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How many odd three-digit integers greater than 800 are there such that 15 Mar 2024, 00:05
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unicornilove
I get 45 options for the 800-899 range..
8(EVEN)X - X has 1,3,5,7,9 = 5 options. 5 EVEN * 5 Options = 25
8(ODD)X - X has 1,3,5,7,9 initially, but since they cant repeat, we will have 4 options for X for every ODD number in tenths place. 5 ODD * 4 Options = 20.

Total is 45.
Bunuel
How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?

A. 40
B. 60
C. 72
D. 81
E. 104


In the range 800 - 899:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)

8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.

Total for this range: \(1*5*8 = 40\).

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is not an option, as it's the first digit)

8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.

Total for this range: \(1*4*8 = 32\).

Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).


Answer: C

Hope it's clear.
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You don't have 5 even options there; you have 4. The digit 8 is already used for the first digit.­
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How many odd three-digit integers greater than 800 are there such that 27 Mar 2025, 09:57
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Hi,

I didn't get the idea of splitting the numbers in 2 subsets:

Though my answer is incorrect but i would like to understand the flaw in my understanding :

Hundreds place could be written in 2c1 ways tens in 9C1 and unit place be filled in 5C1 ways

Is there any way of reaching the answer without splitting the entire numbers ?

walker
Let's xyz is our integer.

x e {8,9}
y e {0...9}
z e {1,3,5,7,9}

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
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How many odd three-digit integers greater than 800 are there such that 27 Mar 2025, 15:26
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MuditKapoor
Hi,

I didn't get the idea of splitting the numbers in 2 subsets:

Though my answer is incorrect but i would like to understand the flaw in my understanding :

Hundreds place could be written in 2c1 ways tens in 9C1 and unit place be filled in 5C1 ways

Is there any way of reaching the answer without splitting the entire numbers ?
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All the digits, in the three digit no. should be distinct

When you take 2C1, 9C1, and 5C1 for hundreds, tens, and units digits respectively, it includes cases such as 919, 877, 955, etc. where 2 digits are similar => hence this approach is incorrect.

Since you know that hundreds digit can either be 8, or 9, we can take each case separately.

8 _ 1 => 8 possible ways to fill tens digit
8 _ 3 => 8 possible ways to fill tens digit
8 _ 5 => 8 possible ways to fill tens digit
8 _ 7 => 8 possible ways to fill tens digit
8 _ 9 => 8 possible ways to fill tens digit

9 _ 1 => 8 possible ways to fill tens digit
9 _ 3 => 8 possible ways to fill tens digit
9 _ 5 => 8 possible ways to fill tens digit
9 _ 7 => 8 possible ways to fill tens digit
You cannot take 9 _ 9, since the digits have to be distinct

Thus, from above 8*9 = 72 possible odd integers between 800 and 1000, having each digit distinct.
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How many odd three-digit integers greater than 800 are there such that 28 Dec 2025, 04:23
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I think we can do it this way;

2*9*4=72, hence C
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How many odd three-digit integers greater than 800 are there such that 25 May 2026, 18:51
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Using the same technique as learned in previous concept by @Bunuel;
Three digits odd integer with non repeating digits above 800
THus, possible range
800-899
900-999
For first range, first digit is 8, thus leaving two digits permutation from the possible range of single digit except 8 (i:e 0,1,2,3,4,5,6,7,9), order is important thus, possible ways of unique arrangements 9P2.
With same logic, for 900 range possible uniqe arrangements 9P2 (except digit "9", ordered combination of two digits from list of other 9 number)

Total possible number combination of unique digits both odd and even (2* 9P2), and we need only odds there is exactly half of it as odd thus,
total number of odds three digits number above 800 is given by (2*9P2)/2=72

Bunuel, is this approach right? Can it be used in all such cases?
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How many odd three-digit integers greater than 800 are there such that 25 May 2026, 21:40
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bishal128
Using the same technique as learned in previous concept by @Bunuel;
Three digits odd integer with non repeating digits above 800
THus, possible range
800-899
900-999
For first range, first digit is 8, thus leaving two digits permutation from the possible range of single digit except 8 (i:e 0,1,2,3,4,5,6,7,9), order is important thus, possible ways of unique arrangements 9P2.
With same logic, for 900 range possible uniqe arrangements 9P2 (except digit "9", ordered combination of two digits from list of other 9 number)

Total possible number combination of unique digits both odd and even (2* 9P2), and we need only odds there is exactly half of it as odd thus,
total number of odds three digits number above 800 is given by (2*9P2)/2=72

Bunuel, is this approach right? Can it be used in all such cases?
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The final numerical answer happens to be correct, but the approach is not correct. You cannot simply count all three-digit numbers with different digits and divide by 2, because odd and even cases are not necessarily split equally. In fact, there are 40 odd numbers from 800 to 899 but only 32 odd numbers from 900 to 999, so if either range were considered alone, this method would give the wrong result. The safe method is to count the units digit directly: 5 * 8 + 4 * 8 = 40 + 32 = 72.
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