Question

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

Answer 1

greaaaaat so nobody knows:(

Answer 2

just looked at it now, thinking

Answer 3

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?

Answer 4

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?

Answer 5

Oh woops, went one exponent too far.

Six-digit numbers don't qualify.

so 1 + 5 + 5^2 + 5^3 + 5^4 = 781

Answer 6

but 5 is an odd number too :)

Answer 7

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?

Answer 8

not sure, but I think not

Answer 9

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

Answer 10

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