Two numbers,when divided by a certain divisor | vedclass

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(B) Let the two numbers be $N_1$ and $N_2$,and the divisor be $d$.According to the problem,$N_1 = q_1d + 43$ and $N_2 = q_2d + 37$,where $q_1$ and $q_

Vedclass · Accepted Answer

$67$

Vedclass · Answer

(B) Let the two numbers be $N_1$ and $N_2$,and the divisor be $d$.According to the problem,$N_1 = q_1d + 43$ and $N_2 = q_2d + 37$,where $q_1$ and $q_2$ are quotients.The sum of the numbers is $N_1 + N_2 = (q_1 + q_2)d + (43 + 37) = (q_1 + q_2)d + 80$.When the sum is divided by $d$,the remainder is $13$. Thus,$80$ divided by $d$ must leave a remainder of $13$.This implies $80 = kd + 13$ for some integer $k$.$kd = 80 - 13 = 67$.Since $67$ is a prime number,the divisor $d$ must be $67$ (as the remainder $13$ must be less than the divisor $d$,and $13 < 67$ holds true).

Two numbers,when divided by a certain divisor,leave remainders $43$ and $37$ respectively. When their sum is divided by the same divisor,the remainder is $13$. Find the divisor.

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