Problem Solving (PS) | Re: After the division of a number successively by 3, 4 and 7

Given that When positive integer x is divided successively by 3, 4, and 7, the remainders are 2, 1, and 4, respectively and we need to find the remainder when x is divided by 84?

We need to solve this problembackwards

- Something divided by 7 and gives 1 as remainder.
- Take that number as quotient for division by 4 and remainder as 1.
- Take the new number formed as the quotient for division by 3 and remainder as 2 to get the value of x.
- Then divide x by 84 to check theremainder.

When an integer (lets say A) is divided by 7, the remainder is4

Theory: Dividend = Divisor*Quotient +Remainder

A -> Dividend
7 -> Divisor
t -> Quotient (Assume)
4 -> Remainder
=> A = 7*t + 4 = 7t +4

When an integer (Lets say "B") is divided by 4, gives A as quotient and 1 asremainder

=> B = 4A + 1 = 4*(7t + 4) + 1 = 28t + 16 + 1 = 28t +17

When x is divided by 3, gives B as quotient and 2 asremainder

=> x = 3B + 2 = 3*(28t + 17) + 2 = 84t + 51 + 2 = 84t + 53

=>
...

Statistics : Posted by BrushMyQuant • on 22 Aug 2021, 02:44 • Replies 3 • Views 1155

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