Factors And Factorials

The key to this is the greatest integer function. Since 7 is the highest prime factor of 42 ,we have to find the following:

Floor of 1000/ $7^1$ = 142

Floor of 1000/ $7^2$ = 20

Floor of 1000/ $7^3$ = 2

Therefore the answer is

142 + 20 + 2 = $\boxed{164}$

$42=2 \cdot 3 \cdot 7$ . However, in $1000!$ , there will obviously be more $2$ 's and $3$ 's than $7$ 's, so all we need to bother about is the number of times $7$ is multiplied in $1000!$ . We simply do $\left \lfloor \frac{1000}{7} \right \rfloor + \left \lfloor \frac{1000}{7^2} \right \rfloor + \left \lfloor \frac{1000}{7^3} \right \rfloor=142+20+2=\boxed{164}$

$\sum _{ i=1 }^{ \infty }{ \left\lfloor \frac { 1000 }{ { 7 }^{ i } } \right\rfloor } =164$

Was also the 164th person to solve the problem :)

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Let $e_{p}(n)$ be the Legendre's Formula for the number $n$ in relation to the prime number $p$ . As we know, we may write $42^{N} = 2^{N} \times 3^{N} \times 7^{N}$ . Then $N$ will be writen as

$N = min (e_{2}(1000), e_{3}(1000), e_{7}(1000))$

But $e_{2}(1000) = 994$ , $e_{3}(1000) = 498$ , $e_{7}(1000) = 164$ . Then $N = 164$ .

as we know prime factors of 42 are 2, 3,7, now in 1000! 7 comes less no. of times comparing to 2 and 3 now for no.of 7 floor of 1000/7=142 floor of 1000/49=20 floor of 1000/343=2 ans=164

use of greatest integer function:

42 = 2 X 3 X 7 the number of times 7 comes in 1000! will determine the greatest power of 42 that can divide 1000! so

1000/7 = 142 (floor of 1000/7 or rounding it of to the lower integer) similarly, 1000/49 =20 and 1000/343 = 2

so 142 + 20 + 2 = 164

This can be solved using the greatest integer function (GIF). Finding the highest power of 42 in 1000! means that we have to find out the highest power of 6 and 7 together in 1000! as they are the prime factors of 42. Note:- these brackets [] indicate the application of the GIF. For eg. [2.236]=2, [2.999]=2 To find out the highest power of 7:- [1000/7]=142 [1000/49]=20 [1000/343]=2 Therefore the highest power of 7 in 1000! is 142+20+2=164

To find out the highest power of 6:- [1000/6]=166 [1000/36]=27 [1000/216]=4 Therefore the highest power of 6 in 1000! is 166+27+4=197

We have to find out the combined power of 6 & 7. Here, the power of 6 exceeds the power of 7. therefore 164 is the highest power of 6x7=42 Therefore the highest power of 42 in 1000! is the power of 7 which is 164.

42 = 7 * 2 * 3 The highest prime factor of 42 is 7. The highest power of 42^N that divides 1000! is the highest power of 7^N.

N = step function(1000/7) + step function(1000/49) + step function(1000/343) = 164