Odd Numbers Starting with Even Number ≤ 100,000

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The discussion focuses on determining how many odd numbers less than 100,000 start with an even digit. It is established that there are 50,000 odd numbers in total below this threshold. The analysis breaks down the problem by examining each range of numbers starting with even digits (2, 4, 6, 8) and calculates the odd numbers within those ranges. The total count of odd numbers starting with an even digit is calculated to be 22,220. This methodical approach highlights the importance of breaking down the problem into manageable parts for accurate results.
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I have a quick question. How many number less than 100,000 are odd numbers but START with an even number??
I was thinking there are 50,000 odd numbers less than 100,000 but how many of those start with an even number?

thanks
 
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Take it one digit at a time. Start within the one's digit. 1, 2, 3, 4, 5, 6, 7, 8, 9.. HOw many of those are even. Then move on to the 10's digit, 10 - 99 how many of those are even, and do you see a pattern between the one's digit place, and the 10's digit place, if so can you extend that to the 100's? 1,000's? 10,000's?
 
blumfeld0 said:
I have a quick question. How many number less than 100,000 are odd numbers but START with an even number??
I was thinking there are 50,000 odd numbers less than 100,000 but how many of those start with an even number?

thanks

not sure if you have learned how to use AP to tackle this qns.

2 4 6 8

For numbers starting with 2, we have 2, (20-29) , (200-299), (2000-2999), (20000-29999)

same for 4 6 and 8

For odd # starting with 2 :

For 2,
0

For 20-29,
1st odd term is 21 , d is 2, last term is 29
Tn = a + (n -1 )d
29 = 21 + (n-1)2
n = 5

For 200 to 299,
1st odd term is 201 , d is 2, last term is 299
Tn = a + (n -1 )d
299 = 201 + (n-1)2
n = 50

For 2000 to 2999,
1st odd term is 2001 , d is 2, last term is 2999
Tn = a + (n -1 )d
2999 = 2001 + (n-1)2
n = 500

For 20000 to 29999,
1st odd term is 20001 , d is 2, last term is 29999
Tn = a + (n -1 )d
29999 = 20001 + (n-1)2
n = 5000

Total # of terms = 4 x (5 + 50 + 500 + 5000) = 22220

I hope this will be of some help
 
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