It’s a question from combinatorial analysis

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Any thoughts? Hint…you know what the last digit is.

– lulu

2 days ago

1

The last digit must be 5

– user381154

2 days ago

Ok, so first count the ones with exactly 44 digits.

– lulu

2 days ago

The ones with 4 digtis are 5³ * 1 = 125

– user381154

2 days ago

Ohh, got it… now i have to multiply 125 by 6 which are the amount of digits that can be in the first spot (5 odd + 1 for the zero) Thanks!

– user381154

2 days ago

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2 Answers

2

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The set of numbers will be

1000 to 99995

-@-−–$-#

Where # could be 5 , so only 1 way to fill that place.

Now @ , it could be 0 or 1 3 5 7 9 , mean 6 ways to fill that place,

$ will be having 1 3 5 7 9 as possible values. So 5 ways.

So the possible numbers

6*5*5*5*1 = 750

It’s 6 * 5 * 5 * 5 * 1

_ * _ * _ * _ * _

Starting from the last digit it can only be “5” so let’s put the 1 on the last spot :

_ * _ * _ * _ * 1

Now let’s focus on counting the ones with exacly 4 digits:

The other 3 digits can only be odd numbers so (1,3,5,7,9) which are 5 possiblities on each spot

_ * 5 * 5 * 5 * 1

And last we need to see how many digits can be on the first spot so :

It can be all the odd numbers (5) and 0 (+1) if it’s smaller then 10000 so the first one is a 6 :

6 * 5 * 5 * 5 * 1 = 750

This is good, though I would have phrased it slightly differently in order to allow for a more easy generalization. Had the question been to find the numbers with those properties lying instead in the range 10,000 to 1,000,000 you wouldn’t have been able to use the six as you had. That would instead have had the answer 53+54+555^3+5^4+5^5. It suits your exact situation though so it gets the job done and does so correctly.

– JMoravitz

2 days ago