Question

5. Find the least 5-digit number which is exactly divisible by 20, 25, 30.

☞ answer the question with full and proper method and explanation...​

Answer 1

Step-by-step explanation:

So, LCM of 20, 25 and 30 is 300. But we want the least 5 digit number, which is exactly divisible by 20, 25 and 30. Least 5 digit number = 10000.

Answer 2

$\large\underline{\sf{Solution-}}$

Since, we have to find the smallest 5 digit number divisible by 20, 25, 30.

So, first we take out the LCM of 20, 25 and 30 using prime factorization method.

$\rm :\longmapsto\:Prime \: factors \: of \: 20 = 2 \times 2 \times 5 = {2}^{2} \times 5$

$\rm :\longmapsto\:Prime \: factors \: of \: 25 = 5\times 5 = {5}^{2}$

$\rm :\longmapsto\:Prime \: factors \: of \: 30 = 2 \times 3\times 5$

So,

$\rm :\longmapsto\:LCM \: ( 20, 25, 30 ) = {2}^{2} \times 3 \times {5}^{2} = 300$

Now,

We know, smallest 5 digit number is 10000.

But we have to find that smallest 5 digit number which is exactly divisible by 20, 25 and 30.

So, using Long Division Method,

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:34 \:\:}}}\\ {\underline{\sf{300}}}& {\sf{\:\:10000 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \: 900 \: \: \:   \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: 1000 \:\:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: 1020 \:  \:\:}} \\ {\underline{\sf{}}}& {\sf{\: \: \: \: \: - \:20  \:\:}}\end{array}\end{gathered}$

So, it means 10000 + 20 = 10020 is the smallest 5 digit natural number divisible by 20, 25, 30

Additional Information :-

Let's take one more example of same type.

Question :- Find the least 5-digit number which is exactly divisible by 20, 25, 30

Solution :-

Since, we have to find the largest 5 digit number divisible by 20, 25, 30.

So, first we take out the LCM of 20, 25 and 30 using prime factorization method.

So,

$\rm :\longmapsto\:Prime \: factors \: of \: 20 = 2 \times 2 \times 5 = {2}^{2} \times 5$

$\rm :\longmapsto\:Prime \: factors \: of \: 25 = 5\times 5 = {5}^{2}$

$\rm :\longmapsto\:Prime \: factors \: of \: 30 = 2 \times 3\times 5$

So,

$\rm :\longmapsto\:LCM \: ( 20, 25, 30 ) = {2}^{2} \times 3 \times {5}^{2} = 300$

Now,

We know, largest 5 digit number is 99999.

But we have to find that largest 5 digit number which is exactly divisible by 20, 25 and 30.

So, using Long Division Method,

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:333 \:\:}}}\\ {\underline{\sf{300}}}& {\sf{\:\:99999 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: 900 \: \: \:   \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: 999 \:\:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: 900 \:  \:\:}}\\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: 999 \:\:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: 900 \:  \:\:}} \\ {\underline{\sf{}}}& {\sf{\: \: \: \: 99 \:\:}}\end{array}\end{gathered}$

So, it means 99999 - 99 = 99900 is the largest 5 digit natural number divisible by 20, 25, 30