First there was this question:

Q. Find the number of numbers from 100 to 400 which are divisible by either 2, 3, 5, and 7

And the obvious solution was to find

  1. numbers divisible only by 2 i.e 100 102 104 ... 400
  2. numbers divisible only by 3 Now we are left with only odd numbers i.e 101 103 105 ... 399 we'll check for only multiples of 3 here, because even multiples of 3 have been taken cared by 2 the numbers will be 105 111 ... 399
  3. numbers divisible only by 5 The numbers left will be 101 103 107 109 113 115 ... 395 We'll check for only multiples of 5 115 125 ... 395
  4. numbers divisible only by 7 Now the numbers left we'll be 101 103 ... 397 And once again we'll check for only multiples of 7

The sum of multiples individually checked in 1, 2, 3, and 4 will give the answer

But then came this question

Q. Find the number of numbers from 100 to 400 which are divisible by any one of 2, 3, 5, or 7

I can't seem to understand is there any difference between the first and the second question, and if there is, then wont the second question have a different answer for each of 2, 3, 5, and 7

Example question by the author, followed by the first question

Book Page

  • 2
    $\begingroup$ The first means "exactly one of" and the second means "at least one of." $\endgroup$ Jun 5, 2012 at 17:45
  • $\begingroup$ @QiaochuYuan Sorry but still I am not getting what you mean by "at least one of", can you give an example? $\endgroup$ Jun 5, 2012 at 17:47
  • $\begingroup$ It means numbers which are divisible by $2$, or $3$, or $5$, or $7$ (not the exclusive or). Thus you also include numbers divisible by $2$ and $3$, and numbers divisible by $3$ and $7$, and... $\endgroup$ Jun 5, 2012 at 17:50
  • $\begingroup$ Off-topic, but regarding your obvious solution: if you simply add up the numbers in 1., 2., 3., and 4., you'll count some numbers more than once. For example, $210 = 2 \times 3 \times 5 \times 7$ gets counted four times (once in 1., once in 2., once in 3., and once in 4.). $\endgroup$
    – talmid
    Jun 5, 2012 at 17:50
  • 2
    $\begingroup$ I don't know if this is helpful, but I asked the question on english.SE: english.stackexchange.com/questions/70066/… My problem with the whole thing is that the either a, b, c, and d doesn't make sense. $\endgroup$
    – Thomas
    Jun 5, 2012 at 18:49

5 Answers 5


This question shows clearly why English is not the best language for mathematics. (French, for instance, is much superior.) I’m a native speaker of the language, and for the life of me, I’m not sure which meaning in the original question is which. A far better way of posing the question would have been “find the numbers divisible by precisely one of 2, 3, 5, and 7”; and “find the numbers divisible by at least one of 2, 3, 5, and 7”.

Another lesson to take from this conundrum is that no matter how clearly you think you’re posing a mathematical question or giving a mathematical explanation, there’ll be someone who misunderstands. Remember too that the author of a text is a fallible human being, often even more fallible than you.


The word "either" typically denotes exclusivity of a logical disjunction. In other words, given statements $x_1,...,x_n$, the following are equivalent:

(1) Either $x_1$ or ... or $x_n$ holds.

(2) Exactly one of the statements $x_1,...,x_n$ holds.

(3) (At least / No less than) one of the statements $x_1,...,x_n$ holds, and (at most / no more than) one of them holds.

Unfortunately, the word "either" is often misused, even by people (e.g.: English-speaking mathematicians) who really ought to know better! Some use it simply as a lead-off word for a list of options, and in that case, it is effectively a throwaway word, meaning nothing at all, and so allowing that any non-null subcollection of the list of statements may hold simultaneously.

As a further complication, the phrase "any one of" seems to carry with it the connotation of "exactly one of"! It would be better to say "any of" instead--as that suggests that more than one may hold, but at least one must hold--but again, this is a common error, even among those who should know better.

Ultimately, this is just a frustrating reality. The best advice I can offer is to try to determine which meaning is intended by the questioner. If it is a textbook, try to find an early example where "either" is used, and see if it means "exactly" or if it is a throwaway. Likewise with "any one of". It is probable that they are intended to have different meanings, so if one is exclusive, then the other is almost surely inclusive.

Edit: Based on what you've told me, it looks as though the author is using "either" as a throwaway word, which suggests that "any one of" is intended to mean "exactly one of" in this text.

Your approach to the first problem is then the right one to take. I am going to break things down as follows, to give a slightly more general approach that should help out in both problems and more, if you're familiar with the basic concepts of set theory (specifically: union, intersection, complement, universe of discourse) and how we may use venn diagrams to help sort out the elements (or cardinalities of sets) of a universe of discourse.

4-category Venn diagram

Here, the universe of discourse ($U$) will be taken to be the integers from $100$ to $400$ (inclusive). $A$ will be taken to be the set those of elements of $U$ that are divisible by $2$; $B$, by $3$; $C$, by $5$; $D$, by $7$.

First, we will determine the cardinalities of (number of elements in) each of the sets $A,B,C,D$ (using the technique of common differences described by the author): $$|A|=151,\: |B|=100,\: |C|=61,\: |D|=43.$$ Next, we determine the cardinalities of the intersections of any pair of the sets $A,B,C,D$ (again with common differences): $$|A\cap B|=50,\: |A\cap C|=31,\: |A\cap D|=21,\: |B\cap C|=20,\: |B\cap D|=15,\: |C\cap D|=9.$$ Then we look at intersections of any trio of the sets $A,B,C,D$, similarly: $$|A\cap B\cap C|=10,\: |A\cap B\cap D|=7,\: |A\cap C\cap D|=4,\: |B\cap C\cap D|=3.$$ Now the intersection of all of them has only $210$ as an element, so $|A\cap B\cap C\cap D|=1$. Thus, there is only one number on the list divisible by all four of $2,3,5,7$. To find how many are divisible by exactly three, we take those divisible by at least three (in this case, members of intersections of trios) and toss out the one divisible by all four. In particular: $$|A\cap B\cap C\cap(\neg D)|=9,$$ $$|A\cap B\cap(\neg C)\cap D|=6,$$ $$|A\cap(\neg B)\cap C\cap D|=3,$$ $$|(\neg A)\cap B\cap C\cap D|=2,$$ so there are a total of $9+6+3+2=20$ numbers on the list divisible by exactly three of $2,3,5,7$.

stage 2 of Venn diagram

To determine how many are divisible by exactly two, we take those divisible by at least two (members of intersections of pairs) and toss out those divisible by all four and those divisible by exactly three. For example, $|A\cap B\cap(\neg C)\cap(\neg D)|=50-1-9-6=34$, and we similarly find $$|A\cap(\neg B)\cap C\cap(\neg D)|=18,$$ $$|(\neg A)\cap B\cap C\cap(\neg D)|=8,$$ $$|A\cap(\neg B)\cap(\neg C)\cap D|=11,$$ $$|(\neg A)\cap B\cap(\neg C)\cap D|=6,$$ $$|(\neg A)\cap(\neg B)\cap C\cap D|=3,$$ giving us a total of $80$ numbers on the list divisible by exactly two of $2,3,5,7$.

stage 3 of Venn diagram

For those divisible by exactly one, start with those divisible by at least one, and toss those divisible by more. So, $|A\cap(\neg B)\cap(\neg C)\cap(\neg D)|=151-1-6-3-9-34-18-11=69$, and simliarly, $$|(\neg A)\cap B\cap(\neg C)\cap(\neg D)|=34,$$ $$|(\neg A)\cap(\neg B)\cap C\cap(\neg D)|=17,$$ $$|(\neg A)\cap(\neg B)\cap(\neg C)\cap D|=11,$$ giving us $131$ numbers on the list divisible by exactly one of $2,3,5,7$. Unless I made a mistake, $131$ should be the answer to the second problem you encountered, if "either" and "any one of" are, in fact, intended to mean different things and "either" is a throwaway word. In total, then, there are $1+20+80+131=232$ numbers on the list that are divisible by at least one of $2,3,5,7$, and since there were only $301$ on the list in the first place, then there are $69$ that are divisible by none of $2,3,5,7$.

final stage

While time-consuming, the advantage of proceeding as above is that, at this point, you can answer any question along this line, since everything is nicely sorted out.

  • $\begingroup$ I've added the page from the book, it also contains an example by the author, see if that helps in understanding the language of his question $\endgroup$ Jun 5, 2012 at 18:32
  • $\begingroup$ Unfortunately, it doesn't help, as it isn't an example of the author's use of "either". Use of "and" and "not" are fairly consistent across the board, but use of "or" and "either" may vary from source to source. $\endgroup$ Jun 5, 2012 at 19:00
  • $\begingroup$ The problem is, there is a forum for guys like us who are preparing for an exam from this book, there the answer to the 1st question includes numbers which are multiples of both 2 and 3 pagalguy.com/forum/quantitative-questions-and-answers/… , but answers here suggest that shouldn't be done, that's why I am getting confused. $\endgroup$ Jun 5, 2012 at 19:02
  • $\begingroup$ I just noticed that the exercise says "either 2, 3, 5, and 7"...which doesn't make any sense at all. What text is this from (so I can recommend against it if it ever comes up)? $\endgroup$ Jun 5, 2012 at 19:02
  • 1
    $\begingroup$ People at english.stackexchange.com/questions/70066/… do agree with the author, atleast one of them $\endgroup$ Jun 5, 2012 at 19:25

I know that there is going to be some disagreements on this, but based on the answer to this question on English.SE, I would say that

Q. Find the number of numbers from 100 to 400 which are divisible
   by either 2, 3, 5, and 7

to mean the number of numbers between 100 and 400 such that the number is divisible by either 2, 3, 5, or 7. The either ... and seems to be a typo. Anyway, the or usually means in math that you can have one or some, but since we have either, then maybe the best way to read this is and an exclusive or. So you would not count $102$ in the list of numbers since it is divisible by both $2$ and $3$

So then,

 Q. Find the number of numbers from 100 to 400 which are divisible 
 by any one of 2, 3, 5, or 7

I would take as the number of numbers that are divisible by any of them. So here you would count $102$ because this is divisible by $2$ (or your argument could be to count $102$ because it is divisible by $3$).

NOTE: I know that this is probably not the "correct answer" in the book. At the end of the day, the author has made some mistakes. So I would agree with Cameron when he in his answer says that you probably want to read through the book and see if you can figure out what the author means when he uses for example "either ... or".

  • $\begingroup$ I totally agree with you.I guess the author has made some mistakes, and if this question comes up on exam, I guess I'll have to leave it, since it has a lot of misconceptions :) $\endgroup$ Jun 5, 2012 at 19:45
  • $\begingroup$ Yeah... you might want to see if you can figure out from you book how the author uses the different terms. But hopefully on your exam things will be clearer. $\endgroup$
    – Thomas
    Jun 5, 2012 at 19:46

I believe the second question is asking for the numbers which are multiples of any of the given numbers, whereas the first question is asking for numbers divisible by only one of the numbers given.

Mathematically, if we let $S_{n}$ be the set of numbers between 100 and 400 which are divisible by n, then we can write the first question as:

$$\#(S_{2}\cup S_{3} \cup S_{5} \cup S_{7})$$

Whilst the second question can be written as:

$$\#(S_{2} \cup S_{3} \cup S_{5} \cup S_{7})-\#((S_{2}\cap S_{3})\cup(S_{2}\cap S_{5})\cup(S_{2}\cap S_{7})\cup(S_{3}\cap S_{5})\cup(S_{3}\cap S_{7})\cup(S_{5}\cap S_{7}))$$

In algorithmic terms, to get the answer to the questions, you would compute the size of the set of numbers divisible by $2$ or $3$ or $5$ or $7$ as you did, to get the answer to the first question. The second question can be found simply by testing if each number is a multiple of only $2$ or $3$ or $5$ or $7$, if it is, then add one to your count; this will give you the correct answers.

Hope this helps.

  • $\begingroup$ won't both of them give the same answer? this is what is confusing me $\endgroup$ Jun 5, 2012 at 18:12
  • $\begingroup$ No, the first case will exclude any numbers which are multiples of both, for instance, $2$ and $5$, whereas the second will include those numbers exactly once. $\endgroup$ Jun 5, 2012 at 18:13
  • $\begingroup$ In the first case, while including even numbers, I am including 100 which is a multiple of both 2 and 5 $\endgroup$ Jun 5, 2012 at 18:15
  • $\begingroup$ @KartikAnand to answer the first question, you should exclude any numbers which are multiples of more than one of the divisors. So $100$ for instance, should be excluded from the set. $\endgroup$ Jun 5, 2012 at 18:16
  • $\begingroup$ So for example while finding multiples of 2, I would only select powers of 2, and multiples of 2 with primes other than 3, 5, and 7 right? $\endgroup$ Jun 5, 2012 at 18:20

In the absence of the first question the second would probably be understood to have the same meaning, which could be answered by (1) compiling a list of the numbers in the range 100-400 that includes all of the numbers divisible by 2 plus, of the remaining numbers, all that are divisible by 3, plus, of the remaining, those divisible by 5, plus, of the remainder, the few that are divisible by 7, and then (2) counting the numbers in that compiled list.

However, the challenge to interpret the second differently would probably be answered by figuring out how many numbers in this range are divisible by 2, 3, 5, and 7--no matter which you choose. Because all four of these divisors are prime numbers, the answer is just one: the product of 2 times 3 times 5 times 7, or 210. There is no smaller number, and the next in the series would be 420, beyond the 100-400 range.

All in all, a nice pair of exercises, despite the linguistic difficulties.


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