Question

On dividing 2121 as well as 624 by a 3 digit number N, we get the same remainder. What is the sum of the digits of N?​

Answer 1

$\large\underline{\text{Answer}}$

22

$\large\underline{\text{Note}}$

The symbol '$\cdot$' stands for the product of numbers.

$\large\underline{\text{Solution}}$

We know that this equation is always true,

$\implies A=BQ+R$

which is called the division algorithm. And,

• $A$ is the dividend.
• $B$ is the divisor.
• $Q$ is the quotient.
• $R$ is the remainder.

Hence, we get two equations for $N$,

$\implies2121=NA+R$

$\implies624=NB+R$

Subtracting two equations we get,

$\implies 2121-624=NA-NB+R-R$

$\implies1497=(A-B)N$

According to the division algorithm, 1497 is divided by $N$ without any remainder. Hence, $N$ is a factor of 1497.

The factorization of 1497 is $3\cdot499$.

For the factor, $N$ to be a three-digit,

$\implies3^{0}\cdot499^{1}=499$

The required number $N$ is 499. Hence, the sum of the three-digit number is 22.

$\large\underline{\text{Verification}}$

$\implies2121=499\cdot4+125$

$\implies624=499\cdot1+125$

Both numbers have the same remainder, 125. Hence, it is verified that $N$ is indeed 499.

Answer 2

Step-by-step explanation:

given :

• On dividing 2121 as well as 624 by a 3 digit number N, we get the same remainder. What is the sum of the digits of N?

to find :

• What is the sum of the digits of N?

solution :

• A=BQ + R

two number of N:

• 2121 = NA + R

• 624 = NB +R

subtract number :

• 2121 - 624= NA- NB + R-R

• 1497 = (A - B)N

factor of N :

• 3. 499 = 499

• 2121= 499 . 4 + 125

• 624 = 499 .1 + 125

• hence, the number N, we get the same remainder = 499