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# The smallest number which when divided by 4, 6 or 7 leaves a remainder of 2, is (1993)

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The smallest number which when divided by 4, 6 or 7 leaves a remainder of 2, is [1993]

a.86     b.80     c.62     d.44

Solution:

LCM:

This problem can easily be solved if we find out the LCM of 4, 6 7 and add 2 to that. But finding out LCMs is easy when all the numbers are the factors of one the given numbers. For example, The LCM of 4, 12, 6 is 12 because the other two numbers are the factors of 12. If you have a group of numbers where one or two numbers are totally different or two numbers are even and one is an odd number like the one given here. you might take time to find the LCM of these numbers. This is why we will use a trick to do it.

Tricky Maths:

This problem can easily be solved if we find out the LCM of 4, 6  and  7.

The error-free method to solve this type of problem is to first multiply all numbers.

4 x 6 x 7 = 24 x 7  = 168

Now divide 168 by 2

$\frac { 168 }{ 2 } =84$

And further, divide this by 2 to make it the smallest.

$\frac { 84 }{ 2 } =42$

Now check whether it divisible by all the numbers.

In this case, it is not divisible by 4, this means we have to take a greater number than 42 and that is 84 and check again.

Yes, the number 84 is divisible by all the numbers.

Now, we have to add 2 to it so that we can get a remainder 2.

84+2=86

Similarly, if we simply calculate the LCM of 4, 6 and 7 which is 84 and add 2 to it, it becomes 86.

Hence, the option (a) is the correct answer.

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