Question

Find the smallest number that when divided
by 35, 56, 91 leaves the remainders of 7 in
each case.​

Answer 1

AnswEr:-

Your Answers s 3647.

ExplanaTion:-

Here we have to find a number which when divided by 35, 56, 91 leaves remainder of 7.

So For this first of all we have to find a number which is divisible by all the three numbers then we add 7 to that number to get answer.

$\rule{200}1$

ExaMple:-

For example if there is a number 4 when divided by 2 and 4 leaves 0 as remainder but if we add 1 to 4 = 5 then it will leave one as Remainder in both the cases.

$\rule{200}{1}$

So let us find a number which is divisible by 35, 56 and 91 by finding the Lowest Common Multiple (LCM) of the three numbers.

So here,

$\tt\implies35 = 5 \times 7. \\ \\\tt\implies56 = 2 \times 2 \times 2 \times 7. \\ \\ \tt\implies91 = 13 \times 7.$

$\tt \therefore L.C.M =5\times2\times2\times2\times13\times7. \\ \\ \tt\mapsto L.C.M.= 3640.$

So,

$\tt\implies3640 + 7 = 3647.$

$\therefore$ The the smallest number that when divided by 35, 56, 91 leaves the remainder of 7 is 3647.