Odd Numbers Starting with Even Number ≤ 100,000

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SUMMARY

The discussion focuses on calculating the number of odd numbers less than 100,000 that start with an even digit. The analysis reveals that for each even starting digit (2, 4, 6, 8), the total number of odd numbers can be derived using arithmetic progression (AP). Specifically, the calculations show that there are 22,220 odd numbers starting with the digit 2, and similar calculations can be applied for the other even digits, confirming a consistent pattern across the ranges of 20-29, 200-299, 2000-2999, and 20000-29999.

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blumfeld0
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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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