Arithmetic
Divisibility:
Let’s learn further application of this topic.
Scenario 1:
How many positive integers are there less than 1000 inclusive, that are divisible by 2 and 5?
Solution:
Remember that, if you have any issue in large scale valued questions, always try to narrow down the scale and check whether things are correct if you use a specific way. For instance,
We need to find common multiples of both 2 and 5, that exists between 1 and 1000 inclusive (i.e within all positive integers less than 1000 inclusive).
For that purpose, we should simply multiply 2 and 5, which results 10. Now divide (let’s say) 30 by 10, you’ll get 3. Now check whether there are only 3 integers exists between 1 and 30 inclusive that are divisible by both 2 and 5? After a quick review, you’ll find the following numbers:
10, 20 and 30
This shows that we were on the right way, so let’s apply this to the required large scale:
1000⁄10 = 100 Answer
Scenario 2:
How many positive integers are there less than 1000 inclusive, that are divisible by 2 or 5?
Solution:
Now you see that there is only difference of ‘and’ and ‘or’ between the two totally different scenarios. This second scenario is more difficult than the first one.
Here, first you should divide 1000 by 2, to get multiples of 2 that exists between 1 and 1000 inclusive (i.e within all positive integers less than 1000 inclusive):
⇒ 1000⁄2 = 500
Now, divide 1000 by 5, to get multiples of 5 that exists between 1 and 1000 inclusive:
⇒ 1000⁄5 = 200
Note that the correct answer would not be 500 + 200 = 700, because there are some integers between 1 and 1000 that are divisible by both 2 and 5 (as we saw in previous scenario). Now these 100 integers will be counted twice if we simply add 500 and 200, because 100 integers exists in both multiples of 2 and multiples of 5 that are divisible by 10 (i.e divisible by both 2 and 5). But we need to find integers that are divisible by only 2, divisible by only 5 or divisible by both. It’s similar to ‘Venn diagram’ topic that you’ll learned in Quantitative Lecture-10. So to avoid double count, you should add 500 and 200, that results to 700. And then subtract 100 from this for remove the overlapping region once, so that region of both counted only once.
⇒ (500 + 200) – 100 = 700 – 100 = 600 Answer
Scenario 3:
How many positive integers are there less than 1000 inclusive, that are divisible by 2, 5 and 7?
Solution:
As mentioned earlier, in scenario 1, this is easy to solve. Simply divide 1000 by the product of 2, 5 and 7.
⇒ 1000⁄70 = 100⁄7 = 14.abc…
We just need to take integer value. For instance, if you divide 10 by 3, you’ll get 3.abc…. As there are 3 multiples of 3 between 1 and 10 inclusive, so 3 is answer. Similarly, in above case of multiples of 70 between 1 and 1000 inclusive,
⇒ 14 Answer
Scenario 4:
How many positive integers are there less than 1000 inclusive, that are divisible by 2, 5 or 7?
Solution:
This question is among the hardest difficulty level scenario, and you’ll be witness after looking explanation.
Firstly,
Number of integers that are divisible by 2:
1000⁄2 = 500
Secondly,
Number of integers that are divisible by 5:
1000⁄5 = 200
Thirdly,
Number of integers that are divisible by 7:
1000⁄7 = 142.abc… ⇒ 142
Now,
Total = 500 + 200 + 142 = 842
Furthermore,
Number of integers that are divisible by both 2 and 5:
1000⁄10 = 100 (These number of integers come twice in total 842 integers)
Furthermore,
Number of integers that are divisible by both 2 and 7:
1000⁄14 = 71.abc… ⇒ 71 (These number of integers come twice in total 842 integers)
Furthermore,
Number of integers that are divisible by both 5 and 7:
1000⁄35 = 28.abc… ⇒ 28 (These number of integers come twice in total 842 integers)
Now, after subtraction:
842 – (100 + 71 + 28) = 643
Finally,
Number of integers that are divisible by 2, 5 and 7, altogether:
1000⁄70 = 14.abc… ⇒ 14
Now, here comes the hardest point. Note that 14 integers come thrice in 842, because these integers are in multiples of 2, in multiples of 5 as well as in multiple of 7. But don’t make mistake to subtract it twice at this stage, because you have subtracted these integers three times already 🙂
For instance,
These integers come in multiples of (2 and 5), in multiples of (2 and 7) and in multiples of (5 and 7). And these you’ve already been subtracted. But As you were required to subtract only twice, rather than thrice. Therefore you need to add these 14 integers now to include these in final answer. i.e
643 + 14 = 657 Answer
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Scenario 5:
What is the maximum possible value of integer n, if 200! is divisible by 5n ?
Explanation:
Perhaps you know the term factorial in Math. If not, let me help. If P is a positive integer, P! is equals to the product of all positive integers from 1 to P inclusive. For instance, 4! = 1 × 2 × 3 × 4 = 24
Know you have refreshed this concept, let’s solve the question.
According to the given condition:
The process is simple, but conceptual.
First step:
If you divide 200 by 5, you’ll get multiples of 5 that are less than 200 inclusive. And that are 40 integers.
(For imagination: it will be 5, 10, 15, 25, …… and 200)
Second step:
Notice that are are few integers in this multiples of 5 can be breakdown into two different 5s.
In order to find number of integers that can be broken down into two different 5s, you can simply divide 200 by 52, which is equals to 25.
After doing so, we’ll get 8 integers that would have at least two 5s.
(For imagination: it will be 25, 50, 75, …… and 200)
Notice that in first division of 200 by 5, we get 40 integers; this shows 40 integers have one 5 in its prime factorization. These integers have cancelled out one 5 by dividing from 5.
Rough Working for novice students:
Here's some rough working to clear your ambiguity in this: If, let's say, you divide 25 by 51, you'll get 5 integers; this shows that there are 5 integers less than 25 inclusive that have one 5 if broken down into prime factorization. These integers are: 5, 10, 15, 20 and 25 But notice that if we divide 25 by 52, you'll get 1 integer; this shows that there is 1 integer less than 25 inclusive that has two 5s, and that is 25. And if you divide 25 by 53, you wouldn't get integer equals to or greater than 1; this shows there's no integer exists less than 25 inclusive that has three 5s if broken down into prime factorization. (Concept of prime factorization is discussed in FREE Study plan for beginners).
Third step:
Now if you divide 200 by 53, which is equals to 125, you’ll still get answer of 1 point something (more specifically 1.6). This means that there is 1 integer less than 200 inclusive that can be broken down into three 5s, and that is 125.
It is important to note that the integer 125 was existed in multiples of 51 = 5, in multiples of 52 = 25 as well as in multiples of 53 = 125.
But let’s take an inter 125. When it was divided by 5, it one of the three 5s was removed. After that when it was divided by 52, its second 5 also removed. But the last 5 was left over. So when we divided 200 by 53, the last 5 also cancelled out. Therefore,
Total number of 5s in 200! = 40 + 8 + 1 = 49 Answer
Let’s attempt another similar question, but with little bit harder scenario:
Scenario 6:
What is the maximum possible value of integer x, if 200! is divisible by 10x?
This question is very important and has come in exam. Here many students quickly but wrongly answer as 20, which is wrong. And few students do similar calculation as above for value of 10 and find answer as 22, which is unfortunately again wrong. 🙂
The point that you forgot while answering this question is that you have missed to take prime factorization of 10. If you ever read chemistry (which is for few people very boring subject), you might believe that when you mix two substances, their individual molecular bonds tend to collapse, and form new one with the other substance. That’s what numbers do!
10 = 5 × 2
Why we need to do so is because few numbers are only multiple of 5, while few are only multiple of 2. For instance, in 200!, both 15 and 12 come, which are not divisible by 10. So when we breakdown 10 to its primes, there we’ll able to get maximum number of those numbers that are divisible by 2 and 5 individually, rather than combined.
Now, think whether multiples of 5 are more than 2? No! clearly, multiple of 2s are more than multiples of 5s. Therefore, we ignore multiples of 2.
And we’ll divide 200! by 5x, and will find out maximum possible value of x in that case. Which is 49, as we calculated in previous scenario. That will be the same answer as when 200! is divided by 10x.
For instance, if you divide 200! by 2x, it will result to more numbers than if we divide by 5x.
e.g,
200! by 2x will give max value of x = 197
and
200! by 5x will give max value of x = 49
So we can say that there are maximum 49 integers that will cancel out with 2s and 5s raised to power same power i.e (2 × 5)x = 10x
Therefore, maximum possible value of integer x = 49 Answer
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Similarly,
What is the maximum possible value of integer x, if 200! is divisible by 15x?
It will also give same answer, because 15 can be broken down to 3 and 5. Multiples of 5 are less than that of 3, thus 49 will be the correct answer again.
Also if it ask:
What is the maximum possible value of integer x, if 200! is divisible by 35x?
Here we’ll find out multiples of 7x, because 35 = 7 × 5
And 7 has less multiples then 5.
Now, let’s go to further dept.
Scenario 7:
What is the maximum possible value of integer x, if 200! is divisible by 25y?
What is the maximum possible value of integer x, if 200! is divisible by 50y?
In both of the cases (i.e 25y and 50y), answer would be same, because both of the integers have same greatest prime factor (i.e 5 which come twice like: 52)
You should now think smart, rather than do hard. Just divide 200! by 5z, that will give you maximum number of 5s that can be divided (and hence cancelled out) by 200!. Simply divide these numbers (i.e z) by 2. Because this will give us the maximum number of 52s that can be cancelled out with 200! after dividing.
Note that we’ll not divide 200! by only 5s, because for instance, 5, 10, 15, 30, … these are numbers exists in 200! that will be cancelled out by 5 raised to power 1, but we need to cancel out 5 raised to power 2. As there are two greatest primes in prime factorization of 25 and 50.
Therefore,
200! divided by 5z will gives maximum value of z = 49
Now divide z by 2 will give value of y = 24 Answer
Note that we’ve ignored one of the 5s from 49 different 5s. The reason is that this one 5 will not cancel with 52, therefore we have to ignore it.
If you not able to understand this concept, you may fee free to ask me on WhatsApp. Your instructor is always ready to help you.
Let’s go ahead!