## azlink10 4 years ago how many positive odd integers less than 70,000 can be represented using the digits 2,3,4,5,and 6

greaaaaat so nobody knows:(

2. Andras

just looked at it now, thinking

you mean as a arithmetic combination of these numbers or does it mean like you can use only 2,3,4,5,6 to represent the number? like 23 is a valid number but 37 is not because it uses 7?

4. 3pwood

I'm sure there's some fancy statistics way to do this, but thinking about it the long way... For one-digit numbers: only 3 works so 1 For two digit numbers: 3 still has to be the second digit, but any of the 5 numbers can be first, so 5 For three digit numbers: 25, by similar logic for four: 125 for five: 625 for six: 3125 so 1 + 5 + 5^2 + 5^3 + 5^4 + 5^5 = 3906 I'm not totally sure of my method. That make good sense?

5. 3pwood

Oh woops, went one exponent too far. Six-digit numbers don't qualify. so 1 + 5 + 5^2 + 5^3 + 5^4 = 781

6. Andras

but 5 is an odd number too :)

7. 3pwood

Right...somehow didn't even see the 5. Apologies. What would that do?...it would mean that both 3 and 5 could be the last digit, and make for twice as many possibilities, right?

8. Andras

not sure, but I think not

9. Andras

I give up, it is too late here :) tomorrow I will give it another go if nobody solved it till then

10. 3pwood

I'm relatively confident that it's 781*2 = 1562, but I know my work is pretty sloppy.