Question

When number successively divided by 6 and 7 leaves remainder 3 and 5 respectively find last remainder when number divided by 8 and 7

Answer 1

Answer:

3 and 2

Step-by-step explanation:

successively by 6 ,7 remainders 3,5

7+5 =12

12*6 =72

72+3=75

so successively divide by 8 and 7 of 75 we get

remainders as 3 and 2

Answer 2

Answer:

Successive remainders would be 4 & 5 respectively.

Step-by-step explanation:

As a successive division takes place. Then considering the later’s divisive status will bring for a divisor 7,the quotient as m the dividend is $(7m+6)$

So, in case of primary division the number will be $6×(7m+6)+3$

,as the divisor here to be 6, quotient is $(7m+6)$

and remainder 3. So number $N= 6(7m+6)+3= 42m+36+3=42m+39$

Then after division by 7 yields$7(6m+5)+4 =42m+39.$

Here quotient $(6m+5)$

and remainder 4. In successive division by 6 ie for division of $(6m+5)$

by 6 will leave remainder 5. Hence successive remainders would be 4 & 5 respectively.