Question

Find the smallest number which leaves remainders 8 & 12 when divided by 28 & 32 respectively.

Answer 1

Hint: In this question use the Euclid's division lemma which is written as Dividend = divisor multiplied by quotient + remainder.

Complete step-by-step answer:

Firstly we will calculate the smallest number which when divided by 28 and 32 gives remainder 0.

For this we need to find the LCM of 28 and 32 first:
\[28 = {2^2} \times 7\]
$32 = {2^5}$
So, LCM ${2^5} \times 7 = 224$

Now by Euclid's division lemma:
Dividend = divisor multiplied by quotient + remainder
$
  224 = 28 \times 8 + 0 \\
   \Rightarrow 224 = 224 + 0 \\
$
Because we want remainder as 8 therefore we can write above result as:
$ \Rightarrow 224 = 216 + 8$
$
   \Rightarrow 224 = \left( {28 \times 8 + 20} \right) + 8 \\
   \Rightarrow 224 - 20 = 28 \times 7 + 8 \\
$

Consider above equation as first equation
Similarly,
$224 = 32 \times 7 + 0$
$ \Rightarrow 224 = 224 + 0$
Because we want a remainder as 12 therefore we can write above result as:
$ \Rightarrow 224 = 212 + 12$
$
   \Rightarrow 224 = \left( {32 \times 6 + 20} \right) + 12 \\
   \Rightarrow 224 - 20 = 32 \times 6 + 12 \\
$
Consider above equation as second equation
Next step is, now observe both the equation first and second and we conclude that:
LHS of both the equations are equal.
Divisors are 28 & 32, as required.
Remainder are also the required number 8 & 12.
So the smallest dividend $224 - 20 = 204$
Hence 204 is the required answer.

Note: Whenever such a type of question comes try to find the common factors of given numbers for this we need to calculate the LCM after that we will apply the Euclid's division lemma so that we can get the smallest number which when divided by 28 & 32 leaves the remainder 8 & 12.