Submitted by mark alvero

In order to be an odd number, $\frac{10!}{N}$ should not be the multiple of any power of $2$ . So, all the powers of $2$ that are contained it $10!$ must also be contained in $N$ , so that they get cut while dividing and no power of $2$ remains in $\frac{10!}{N}$ . So, on expanding $10!$ , we get: $10! = 1 \times2 \times3 \times4 \times5 \times6 \times7 \times8 \times9 \times10 = 2^8 \times3^4 \times5^2 \times7$ . Since $10!$ has $2^8$ contained in it, $N$ should compulsorily also contain $2^8$ in it. It is not required for any other number to be contained in $N$ in order to make $\frac{10!}{N}$ an odd number, thus the smallest possible value of $N$ is $2^8 = \fbox{256}$ .

First of all, you need to know that any number multiplied with an even number always yields an even number. Therefore, odd * even = even and even * even = even.

Now, lets proceed. We know $\frac{10!}{N}$ can be written as $\frac{10\times9\times8\times...\times1}{N}$ .

In order to yield an odd number, N should be such that it cancels all the even factors in the numerator.

Even numbers in the numerator are 10, 8, 6, 4 and 2. So, N would be $10\times8\times6\times4\times2$ . This would cancel all the even terms in the numerator and we would get an odd result. But, here the smallest value of N is asked. So we further factorize 10, 8, 6, 4, and 2. The odd factors in them are:

$10 = 2\times5$

$8 = 2\times2\times2 = 2^{3}$

$6 = 2\times3$

$4 = 2\times2 = 2^{2}$

$2 = 2^{1}$

Taking all the even factors, we finally have N = $2\times2^{3}\times2\times2^{2}\times2^{1} = \boxed{256}$

That's the answer!

As factorial of 10 is equal to $10\times9\times8\times7\times6\times5\times4\times3\times2\times1$ , we can multiply all the even numbers to get $N$ because the product of an even number wth any other number is even, so if there is no even number, we can get an odd integer

$N = 10\times8\times6\times4\times2 = 3840$

BUT this isn't the answer, because it wants the smallest number and this is a really large number, so maybe we can get a smaller number

Notice that $10 = 5\times 2$ and $6 = 3\times 2$

So we can break them down into $5\times 2\times 9\times 8\times 7\times 3\times 2\times 5\times 4\times 3\times 2\times 1$

Then, we multiply the even numbers altogether to get $N$

$N = 2\times 8\times 2\times 4\times 2$

$N = 256$

Well, according to Legendre, when we decompose $10!$ on prime factors we will get

$n = \left \lfloor \frac{10}{2} \right \rfloor + \left \lfloor \frac{10}{2^{2}} \right \rfloor + \left \lfloor \frac{10}{2^{3}} \right \rfloor = 8$

Prime factors $2$ . Then, we have to devide $10!$ by at least $N = 2^n = 256$ in order to $10!/N$ to be odd.

We know if 10! is 1. 2. 3. 4. 5. 6. 7. 8. 9. 10 = 2^8. 3^4. 5^2. 7. if 3^4. 5^2. 7 is an odd integer, thus 2^8 is smallest value that make 10!/N is an odd integer. Answer : 256.

In order that 10!/N is an odd integer , N should eliminate all ' 2 ' s in the number 10 ! . Counting 2 == 2 4 == 2^2 6 == 2 8 == 2^3 10 == 2 ie total of 2^10 is even rest are odd thus N = 256

For $N$ to be an odd integer, all the $2$ 's must be knocked off from $10!$ . There are $8$ two's in $N$ and hence the number $N$ has to be divided by $2^{8}$ . This gives us an odd number and thus the correct answer is $2^8 = \boxed{256}$ .

10!= 10 9 8 7 6 5 4 3 2 1 = (5 2) 9 8 7 (3 2) 5 4 3 2 1 =5 9 8 7 3 5 4 3 (2 8 2 4 2) = 5 9 8 7 3 5 4 3 256 thus, dividing by 256 would give an odd integer.

Highest power of 2 in 10! is 8 hence N = 2^8=256

10! = 10x9x8x7x6x5x4x3x2x1

10!/(16) =10x9x7x6x5x4x3x1 =======first 16

= 2x5x3x3x7x3x2x5x2x2x3x1

(2x5x3x3x7x3x2x5x2x2x3x1)/(2x2x2x2) ====== second 16

=5x3x3x7x3x5x3

Therefore the smallest number is 16x16=256

10!/N=(10x9x8x7x6x5x4x3x2x1)/N=(5x2)x9x8x7x(2x3)x5x4x3x2x1/N={(2x8x2x4x2)x(5x9x7x3x5x3)}/N (5x9x7x3x5x3) is an odd integer so , smallest value of N is 2x8x2x4x2=256

10!=1 2 3 4 5 6 7 8 9 10. =1 2 3 (2^2) 5 (2 3) 7 (2^3) 9. =(2^8) 1 3 5 3 7 9. if we divide above no. with 2^8 it will become an odd no. i.e the smallest value is 256.