Question

Find the divisors of $ 1980 $ .
i) How many are multiple of $ 11 $ ? Find their sum
ii) How many are divisible by $ 4 $ but not by $ 15 $ .

Answer 1

Hint: In this question we will use prime factorization of $ 1980 $ to find all divisors. Divisors or factors are those numbers which do not leave any remainder upon division that is they divide the number exactly. A prime factor is one which has no further factor except one and the number itself, whereas a composite factor is one which is a combination of several prime factors. Both or a combination of both make up the divisors of a number.

Complete step-by-step answer:
Now, to find the divisors of 1980. We will find all its factors whether prime or composite. Since divisors and factors are the same, as they divide the numbers completely without leaving any remainder.
So to find the factors or divisors we need to first find all the prime factors of $ 1980 $ .
So factorizing $ 1980 $ :
 $ \begin{array}{*{20}{c}}
  2&\hline & {1980} \\
\hline
  2&\hline & {990} \\
\hline
  3&\hline & {495} \\
\hline
  3&\hline & {165} \\
\hline
  5&\hline & {55} \\
\hline
  {11}&\hline & {11} \\
\hline
  {}&\hline & 1
\end{array} $
 $ 1980 = 2 \times 2 \times 3 \times 3 \times 5 \times 11 \times 1 $ .
Since, we have to find all divisors of $ 1980 $ , so all divisors will be including of these prime and the composite factors which will be:
 $
  1 \times 1980 = 1980 \\
  2 \times 990 = 1980 \\
  3 \times 660 = 1980 \\
  4 \times 495 = 1980 \\
  5 \times 396 = 1980 \\
  6 \times 330 = 1980 \\
  9 \times 220 = 1980 \\
10 \times 198 = 1980 \\
11 \times 180 = 1980 \\
12 \times 165 = 1980 \\
15 \times 132 = 1980 \\
18 \times 110 = 1980 \\
20 \times 99 = 1980 \\
22 \times 90 = 1980 \\
33 \times 66 = 1980 \\
36 \times 55 = 1980 \\
44 \times 45 = 1980 \\
  $
So, $ 1980 $ has $ 35 $ divisors. Which are:
\[1,2,3,4,5,6,9,10,11,12,15,18,20,22,30,33,36,44,45,55,66,90,99,110,132,165,180,198,220,330,396,495,660,990{\text{ }}and{\text{ }}1980.\]
Now,
i)To find the number of divisors which are multiples of $ 11 $ ,
Let us first separate out the ones which are multiples of $ 11 $ , and they are:
\[11 + 22 + 33 + 44 + 55 + 66 + 99 + 110 + 132 + 165 + 198 + 220 + 330 + 396 + 495 + 660 + 990 + 1980\]
So, there are a total of $ 18 $ divisors of $ 1980 $ which are multiples of 11.
Their sum will be: \[11 + 22 + 33 + 44 + 55 + 66 + 99 + 110 + 132 + 165 + 198 + 220 + 330 + 396 + 495 + 660 + 990 + 1980\] $
   \Rightarrow 11(1 + 2 + 3 + 4 + 5 + 6 + 9 + 10 + 12 + 15 + 18 + 20 + 30 + 36 + 45 + 60 + 90 + 99 + 180) \\
   \Rightarrow 11(945) \\
   = 10395 \;
  $
Therefore the sum of all divisors of 1980 which are multiples of $ 11 $ is $ 10395 $ .

ii) Next we need to find the divisors which are divisible by $ 4 $ but not by $ 15 $ .
So to find out the divisors which are divisible by 4 but not by 15, let us first separate them:
They will be:
\[44,132,220{\text{ }}and{\text{ }}396\]
Since
 $
  44 = 1 \times 2 \times 2 \times 11 \\
  132 = 1 \times 2 \times 2 \times 3 \times 11 \\
  220 = 1 \times 2 \times 2 \times 5 \times 11 \\
  396 = 1 \times 2 \times 2 \times 3 \times 3 \times 11 \;
  $
So, all these numbers are divisible by $ 4 $ as they have $ 4 $ as their factor but they cannot be divided by $ 15 $ as they do not have 15 as their factor.
Therefore:
There are a total of $ 35 $ divisors of $ 1980 $ .
i) Amongst the $ 35 $ divisors of $ 1980 $ , $ 18 $ divisors are multiples of $ 11 $ .
ii) Out of the $ 35 $ divisors of $ 1980 $ there are $ 4 $ divisors which are divisible by $ 4 $ but not divisible by $ 15 $ .

Note: For bigger numbers, their factors are not necessarily always just the prime factors, some are composite factors as well which are made up of prime factors. Composite factors also divide the numbers exactly just like the prime factors. Instead of applying the prime factorization we can also apply the divisibility rule of $11$ to check which of the divisors multiples are of $11$. According to the divisibility rule of $11$, if the number of digits in the given number is even, we will add the first and subtract the last digit from the rest, if the result must be divisible by $11$, then the number is divisible by $11$. If the number of digits in the given number is odd, then we will subtract the first and last digit from the rest, if the result is divisible by $11$, then the number is divisible by $11$.