M25-20

Question

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

Bunuel · Accepted Answer

Official Solution:

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

pinkcupcake · Answer

Hi Bunuel,

I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit  odd  integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.

Bunuel · Answer

801, 803, 805, 807, ... are all odd integers.

aviram · Answer

Amazingly elegant method.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

joshuaroujdiaz · Answer

I did not quite understand the solution. Hi Bunuel - when figuring out the choices, why didn't we figure out the choices for the second digit first before the 3rd digit? For example,

In the range 900-999
1 choice for first digit: 9
9 choices for second digit: 10 possible digits, except the digit 9 which is the first digit already.
3 choices for third digit: 1,3,5,7,9, but subtract to account for the first and second digits.

Thank you!

Bunuel · Answer

That's not correct because if the second digit selected is even, you won’t have 3 options for the third digit—you’ll have 4. The general rule is to start with the slot that has the most restrictions and then move to the less restricted slots.

Unknown5612 · Answer

Is this correct:
 Total numbers (even and odd): 3*9*8
now from 800 to 999, there will be equal number of odd and even numbers. so total odd numbers: 3*9*8/2=72

Bunuel · Answer

Not really sure 3hat you are doing there but 3*9*8/2 = 108, not 72.

Unknown5612 · Answer

Bunuel ,

Oops typo, Total numbers (even and odd): 2*9*8 (Let ABC be any number, A can be either 8 or 9 - hence 2 possible numbers, B can be 9 numbers and C can be 8 numbers)
now from 800 to 999, there will be equal number of odd and even numbers. so total odd numbers: 2*9*8/2=72

Bunuel · Answer

___________________
Yes, that's correct.

agrasan · Answer

I like the solution - it’s helpful.

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