Please help me with this problem
How many zeroes are there at the end of the expression $4!^{4!}+8!^{8!}+16!^{16!}$
Asked
Active
Viewed 57 times
-2
S.Bansal
- 101
-
Do you have any ideas to start? – BallBoy Jan 12 '18 at 13:41
-
We can star "not sure if correct" by finding the greatest power of 10 in 8! and 16! – S.Bansal Jan 12 '18 at 13:43
-
Split them into factors of 5 and 2. Each pair of 5 and 2 will yield a 0. – prog_SAHIL Jan 12 '18 at 13:44
-
That sounds like a good strategy. Try starting down that path and post again when you see where it takes you. – BallBoy Jan 12 '18 at 13:44
-
I have tried this way but couldn't approach the answer in this way – S.Bansal Jan 12 '18 at 13:45
-
Not a duplicate, but this question could help you : https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n/141197#141197 – Arnaud D. Jan 12 '18 at 16:30
2 Answers
7
Note that $8!$ and $16!$ are divisible by $5$, but $4!$ is not.
Hagen von Eitzen
- 374,180
-
-
@S.Bansal Think about it this way. $8! = 10 \cdot 4032$ Therefore, $8!^n$ has zeros at the end of it. A similar process can be applied to $16!$. However, $4!$ is not divisible by $10$ This means that $4!^n$ is also not divisible by $10$. Note if the sum must be divisible by $10$ (ie have zeros at the end) and two given terms are divisible by $10$ the other must be divisible as well. Since this is not true, there are $0 0's$ left – John Lou Jan 12 '18 at 13:47
5
$8!$ has one zero at its end as $5,2<8$. So there are $8!$ zeros at the end of $(8!)^{8!}$. $5,10,15<16$, so there are three zeros at the end of $16!$. So there are $3\cdot 16!$ zeros at the end of $(16!)^{16!}$. However $(4!)^{4!}$ has no zeros at the end of it. Hence the sum $(4!)^{4!}+(8!)^{8!}+(16!)^{16!}$ has no zeros at the end of it. The last digit of this number is same as the last digit of $24^{24}$, which is $6$.
QED
- 12,644