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In the decimal system of numeration the number of $6$-digit numbers in which the sum of the digits is divisible by $5$ is $$(A)\space180000 \space\space(B) \space540000\space\space (C) \space 5\times10^5\space \space (D) \space \text{none}$$

First, I noted that $a+b+c+d+e+f=5,10,15\dots50$ where $a\geq1$ and $ b, c, d, e, f\geq0$.

From this I got $9C5 + 14C5 +...... 54C5$, which is just too hectic too calculate and I don't know it will give the right answer or not.

Then I observed that there are $900000$ total possible numbers and if I divide it by $5$, I will get the correct answer which is $180000$.

Is it a coincidence or is there any logic to it? What would be the proper method to solve this?

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    $\begingroup$ is 000005 a six-digit number, as that term is intended in the formulation of the question? $\endgroup$ – C. M. Sperberg-McQueen Oct 6 '17 at 18:29
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Hint Suppose you have specified the first five digits of a six-digit number. How many choices are there for the sixth digit such that the sum of all six digits is divisible by $5$?

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Disclaimer: sorry, wrong approach as shown in comments.

It is no coïncidence.

If $n$ is a 6-digits integer and $S(n)$ is the sum of its digits, you will notice that $S(n+1) mod 5 = (S(n) mod 5) +1$

Moreover, as you noticed, the number of 6-digits integers is a multiple of $5$

Hence exactly one fifth of all 6-numbers integers will have a sum of digits that equals $1$ modulo $5$: They are $100000,100005,...,100000+5k,..., 999996$.

One fifth of them will have a sum of digits that equals $2$ modulo $5$, one fifth will have a sum of digits that equals $3$ modulo $5$, one fifth will have a sum of digits that equals $4$ modulo $5$, and finally one fifth will have a sum of digits that equals $0$ modulo $5$, hence be multiples of 5.

They are $100004,100009,...,100004+5k,..., 999995$.

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    $\begingroup$ You're correct that exactly a fifth of all $6$-digit numbers are divisible by $5$, but your reasoning is incorrect. If I set $k=2$ in your latter sequence, then I get $100004 + 10 = 100014$, whose digits sum to $6$, which is not divisible by $5$. @Travis has the correct approach. $\endgroup$ – T. Linnell Oct 6 '17 at 15:23
  • $\begingroup$ Just curious - is there a reason for the ï in coïncidence instead of a regular i? $\endgroup$ – Digital Trauma Oct 6 '17 at 17:14
  • $\begingroup$ @DigitalTrauma it's a coïncidence $\endgroup$ – RiaD Oct 6 '17 at 17:49
  • $\begingroup$ Some people add umlauts to indicate that two vowels are to be pronounced separately instead of together. Without the umlaut (and a first grade education), you might mistakenly pronounce the first syllable in coincidence as "coin". Also not uncommon in cooperate, reentry, etc. In my view it's unnecessary and pretentious. $\endgroup$ – BallpointBen Oct 6 '17 at 17:51
  • $\begingroup$ - The "ï" in "coïncidence" is due to your's truly being French. And French is such a pretentious language that it's how it spells that​ word. If this is my single English mistake in my answer, I will feel very proud about it. - The math mistake pointed by Linnell is much more embarrassing. I'll edit but I stand ashamed. $\endgroup$ – Evargalo Oct 6 '17 at 18:39

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