0
$\begingroup$

$40$-years-old John only knows even digits. How many $4$-digits numbers he can write down? Find the sum of this numbers.

Attempt:

John knows $5$ digits $(0,\ 2,\ 4,\ 6,\ 8)$. We have 4 different places to place them. Certainly there are no $4$-digits numbers that begin with $0$ (that is, we don't take into account, for instance, the number $0846$). So in our first place we can put only $4$ out of out $5$ even numbers. In the remaining places there is no condition imposed, so John can write down:

$$4\cdot5^3 = 500\ \text{numbers}$$

Now let find the sum of these numbers.

A $4$-digit number in the decimal numerical system can be written as:

$$\overline{abcd} = a\cdot 10^3 + b\cdot 10^2 + c\cdot 10^1 + d\cdot 10^0$$

Certainly we want to find the sum of our $500$ numbers so,

$$\sum_{k=1}^{500}\overline{abcd} = \sum_{k=1}^{500}a\cdot 10^3 + \sum_{k=1}^{500}b\cdot 10^2 + \sum_{k=1}^{500}c\cdot 10^1 + \sum_{k=1}^{500} d\cdot 10^0$$

And here I'm stuck. I have to make some conclusions about the coefficient on the RHS, but I don't know which. Does really matter the fact that he is $40$ years old?

$\endgroup$
2
  • 1
    $\begingroup$ $$\sum_{evens}^{}\overline{abcd} = \sum_{p=1}^{4}\left (2p\cdot 10^3 + \sum_{q=0}^{4}\left (2q\cdot 10^2 + \sum_{r=0}^{4}\left (2r\cdot 10^1 + \sum_{s=0}^{4} 2s\cdot 10^0\right )\right )\right )$$ $\endgroup$
    – N74
    Commented Oct 28, 2016 at 6:03
  • 1
    $\begingroup$ You need to change the header. You aren't dealing with all even numbers, only numbers with all digits even. $\endgroup$
    – true blue anil
    Commented Oct 28, 2016 at 8:06

2 Answers 2

Reset to default
2
$\begingroup$

As you have found out, there will be $500$ such numbers.

Except for the thousandth place, the average value of a digit will be $(0+2+4+6+8)/5 = 4,$ while that for the thousandth place will be $(2+4+6+8)/4 = 5$

Thus overall sum $= 500(5\cdot10^3 + 4\cdot10^2 +4\cdot10^1 + 4\cdot10^0) = 2,722,000$

$\endgroup$
1
$\begingroup$

You have a good start and noticed that you can add each tens-place individually. To continue, notice that if we try to add the units digits of each of the numbers:

$$\begin{array}{r}2000\\2002\\2004\\2006\\\vdots\\+8888\\\hline\end{array}$$

We may instead group them according to their last digit:

$$\left(\begin{array}{r}2000\\2010\\2020\\\vdots\\ +8880\\\end{array}\right)\text{all the 0's}\\ \left(\begin{array}{r}2002\\2012\\2022\\\vdots\\ +8882\end{array}\right)\text{all the 2's}\\\vdots~~~~~~~$$

How many times does $0$ appear as the last digit of one of the numbers? How much then does $0$ contribute to the overall sum of the units digit column?

How many times does $2$ appear as the last digit of one of the numbers? How much then does $2$ contribute to the overall sum of the units digit column?

etc...

What then is the total of the units digit column?

Exactly one fifth of the numbers in the list end with $0$. Similarly exactly one fifth of the numbers end with 2,4,6, and 8 respectively. The total sum for the units column of the summation is then $100\cdot 0 + 100\cdot 2+\dots+100\cdot 8 = 2000$

Do so similarly for ten's digit, hundred's digit, and thousand's digit columns.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .