The greatest number which when divides 258 and 323 leaving the remainders 2 and 3 respectively are
A. 32
B. 64
C. 16
D. 128
Answer
619.8k+ views
Hint: We’ll first use the remainder theorem which is $N = pq + r$ where number N is divided by p, then q, and r come out to be the quotient and the remainder respectively, that we’ll apply on the given numbers. From the remainder theorem, we’ll get the 2 multiples of the required number and from those multiples, after finding the HCF of those numbers we’ll approach our answer.
Complete step by step answer:
Given data: 258 and 323 leaves remainder 2 and 3 respectively when divided by the required number.
Let the number be A
The remainder theorem states that if a number N is divided by p, the q and r comes out to be the quotient and the remainder respectively then the relation between them is
i.e. $N = pq + r$
According to the given data when 258 is divided by A leaves remainder 2
Let the quotient be ${q_1}$, so using remainder theorem we get,
$ \Rightarrow 258 = A{q_1} + 2$
$ \Rightarrow 256 = A{q_1}$
And when 323 is divided by A leaves remainder 3
Let the quotient be ${q_2}$, using remainder theorem we get,
$ \Rightarrow 323 = A{q_2} + 3$
$ \Rightarrow 320 = A{q_2}$
Now we can say that the HCF of $A{q_1}$ and $A{q_2}$ will be the greatest value of A
Using the prime factorization method
$ \Rightarrow 256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$ \Rightarrow 320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$
Therefore the HCF will be $2 \times 2 \times 2 \times 2 \times 2 \times 2$ i.e. 64
Therefore the greatest integer value for A will be 64
i.e. (B)64 is the correct option.
Note: Before using the remainder theorem we should acknowledge the terms like divisors dividend quotient remainder.
1) The divisor is the number that divides.
2) The dividend is the number from which the number is being been divided.
3) The quotient is referred as to the number of times the divisor divides the dividend
4) The remainder is referred to as the residual number after the divisor divides the dividend.
Complete step by step answer:
Given data: 258 and 323 leaves remainder 2 and 3 respectively when divided by the required number.
Let the number be A
The remainder theorem states that if a number N is divided by p, the q and r comes out to be the quotient and the remainder respectively then the relation between them is
i.e. $N = pq + r$
According to the given data when 258 is divided by A leaves remainder 2
Let the quotient be ${q_1}$, so using remainder theorem we get,
$ \Rightarrow 258 = A{q_1} + 2$
$ \Rightarrow 256 = A{q_1}$
And when 323 is divided by A leaves remainder 3
Let the quotient be ${q_2}$, using remainder theorem we get,
$ \Rightarrow 323 = A{q_2} + 3$
$ \Rightarrow 320 = A{q_2}$
Now we can say that the HCF of $A{q_1}$ and $A{q_2}$ will be the greatest value of A
Using the prime factorization method
$ \Rightarrow 256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$ \Rightarrow 320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$
Therefore the HCF will be $2 \times 2 \times 2 \times 2 \times 2 \times 2$ i.e. 64
Therefore the greatest integer value for A will be 64
i.e. (B)64 is the correct option.
Note: Before using the remainder theorem we should acknowledge the terms like divisors dividend quotient remainder.
1) The divisor is the number that divides.
2) The dividend is the number from which the number is being been divided.
3) The quotient is referred as to the number of times the divisor divides the dividend
4) The remainder is referred to as the residual number after the divisor divides the dividend.
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