how many positive odd integers less than 70,000 can be represented using the digits 2,3,4,5,and 6
Stacey Warren - Expert brainly.com
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greaaaaat so nobody knows:(
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?
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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
so 1 + 5 + 5^2 + 5^3 + 5^4 + 5^5 = 3906
I'm not totally sure of my method. That make good sense?
Oh woops, went one exponent too far.
Six-digit numbers don't qualify.
so 1 + 5 + 5^2 + 5^3 + 5^4 = 781
but 5 is an odd number too :)
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?
not sure, but I think not
I give up, it is too late here :) tomorrow I will give it another go if nobody solved it till then
I'm relatively confident that it's 781*2 = 1562, but I know my work is pretty sloppy.