Question

The highest power of 7 that completly divides 80! Is :

Answer 1

We know that  80!  is the product of first 80 natural numbers.

80! = 1 × 2 × 3 × ... × 80

In this product we have to find how many '7s' are multiplied!

For this, we separately find out them in such a way that, first no. of multiples of 7 under 80, then that of 7² under 80, then that of 7³ under 80, etc. will be found.

First we find out no. of multiples of 7 under 80.

80 gives quotient 11 on division by 7. So there are 11 multiples of 7 under 80.

Now no. of multiples of 7² = 49 under 80 have to be found.

There is only 1 multiple of 49 under 80, and that is none other than 49, isn't it?

[This multiple of 49 is also added along the multiples of 7, but it's not mentioned.]

Since 7³ = 343 > 80, there's no need of finding no. of multiples of 7³ and other high powers under 80.

Now add the total nos. of multiples, and we get,

11 + 1 = 12

This 12 is the exponent of the highest power of 7 by which 80! has to be divided.

So the power of 7 will be " 7¹² ", won't it be?

Hence 7¹² is the answer.

Answer 2

Answer:

It's very simple.

If we want to find the number which is also the factor of the same number which divides another number then the result can be found simply.

Follow the steps mentioned below:

We know that 7 is a factor of 7.

So from this statement, we come to know that 80 must be divisible by 7 so that we can find that how many times 7 will be multiplied to get 80 .

So now we have to test the divisibility of 80 by 7.

The divisibility process is as follows:

The last digit of 80 is 0.

Putting the divisibility criteria to 80, we get:

2*0=0

8-0=8

We get 8.

But 8 is not divisible by 7.

So 80 is not divisible by 7.

Hence there is no power of 7 which can divide 80 completely.