4 appearing in the integers 1-1000000

**oldguynewstudent** · May 20th 2010, 08:32 PM

You write down all of the integers from 1 to 1,000,000. How many times did you write the digit 4?

I get $\sum_{n=0}^{6}10^{n}$

Is this correct?

**Opalg** · May 21st 2010, 06:45 AM

Originally Posted by oldguynewstudent

You write down all of the integers from 1 to 1,000,000. How many times did you write the digit 4?

One integer in every ten has a 4 in the units position. That gives $10^5$ 4s in the units position (in the integers from 1 to $10^6$ ). Ten integers in every hundred have a 4 in the tens position (in other words, each of the integers ending in 40, 41, ..., 49). That gives $10\times10^4 = 10^5$ (again) 4s in the tens position. And so on. There will be a total of $10^5$ 4s in the hundreds position, in the thousands position, in fact in each of the six positions in a six-digit number. So the total number of 4s that you would have to write, if you wrote down all the integers from 1 to a million would be 600,000.

**oldguynewstudent** · May 21st 2010, 07:25 AM

Thank you so much. As always I was on the right track but off a little.

The neurons are willing but the dendrites are weak!

**Archie Meade** · May 21st 2010, 02:05 PM

$4XXXXX$

X may be any value from 0 to 9.
This counts the number of 4's in the given position from 1 to a million, which is $10^5$

$Y4XXXX$

Y can be any value from 1 to 9, X can be any value from 0 to 9.
However Y can also be missing, hence it has 10 possibilities.

The total number of 4's in this position is $(9)10^4+10^4=(10)10^4=10^5$

$YY4XXX$

Y can be any value from 1 to 9 or missing.
X can be any value from 0 to 9.

The number of 4's in this position is $10^210^3=10^5$

$YYY4XX$

The number of 4's in this position is $10^310^2=10^5$

$YYYY4X$

The number of 4's in this position is $10^410=10^5$

$YYYYY4$

The number of 4's in this position is $10^5$

The total number of 4's is $(6)10^5$

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