​Find the smallest natural number which, when divided by 9,24 and 12 , leaves a remainder 4 in each case but when divided by 5 leaves no remainder.​

Correct option is A

Given:

Given numbers are 9, 24 and 12

Remainder in each case =5

Formula Used:

LCM of the numbers is the smallest common number which when divided by the numbers leaves no remainder

Solution:

Prime factorization of 9,24 and 12

9 = $$3^2$$​​

24=$$2^3 \times 3$$​​

12= $$2^2 \times 3$$​​

LCM = $$2^3 \times 3^2 = 72$$​​

Using Euclid’s division lemma(a= bq +r) we get:

x= 72q + 4   ….(1)

We substitute the value of q and check till we get number divisible by 5

Putting q = 1 in eq(1)

x = 72(1) +4 = 76

76 is not divisible by 5

Putting q = 2 in eq(1)

x= 72(2) +4 = 148

148 is not divisible by 5

When we put q = 3

x= 72(3) + 4  =220

Now 220 is divisible by 5

Hence the number is 220

Alternative Method:

The above method is lengthy hence we use trial method using the options:

Option are 220,144,150 and 148

So the number should be divisible by 5 and leaves remainder 4 when divided by 9,24 and 12

Number 144 and 148 are not divisible by 5 hence we can directly omit them

Number 150 is divisible by 5 but leaves remainder 6 when divided by 9 hence it does not satisfies both condition

Number 220 is divisible by 5 and leaves remainder 4 when divided by 9,24 and 12 hence 220 is the number