Question

The least number which when divided by 5,6,7 leaves the reminder 3,4,5 in each case ?​

Answer 1

Answer:

208

Try to understand this concept:

The remainders are two less than the dividing numbers.

If they are added by 2, what happens to the remainders?

The remainders are equal to the dividing numbers, so there will be no remainder. It lets 5, 6, and 7 divide the number completely.

Step-by-step:

Here the number will be k.

If 2 is added, we get k+2.

5, 6, and 7 leave no remainder for k+2.

→ k+2 is multiple.

→ k+2 needs to be the LCM.

→ k+2=210

From this equation, we obtain 208 as a solution.

Additional Information

Let the number be k.

Then it is $\sf{k\equiv -2\mod 5,\;6,\;7}$

Now it is $\sf{k+2\equiv 0\mod 5,\;6,\;7}$

Therefore $\sf{k+2}$ will be the multiples of 5, 6, 7 because the remainder zero indicates that it is multiple.

Such least number will be the Lowest Common Multiples, or commonly LCM.

Here LCM is 210.

$\sf{k+2=210n\;(n\;is\;any\;natural\;number)}$

$\sf{k=210n-2\;(n\;is\;any\;natural\;number)}$

Therefore, the values such that when divided by 5, 6, and 7 leave the remainder 3, 4, and 5 is 208, 418, 628, ….

Answer 2

Answer:

$\huge\red{ \fbox{anwer}}</p><p>$

208

$\red{ \fbox{concept used}}</p><p>$

The remainders are two less than the dividing numbers.

If they are added by 2, what happens to the remainders?

The remainders are equal to the dividing numbers, so there will be no remainder. It lets 5, 6, and 7 divide the number completely.

$\huge\red{ \fbox{Explanation}}</p><p>$

Here the number will be k.

If 2 is added, we get k+2.

5, 6, and 7 leave no remainder for k+2.

→ k+2 is multiple.

→ k+2 needs to be the LCM.

→ k+2=210