benatudq
2022-10-23
Answered

What is the digit in the ten's place of ${23}^{41}\ast {25}^{40}$? How do you calculate this? The usual method for this kind of problem is using the Binomial theorem.

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DoryErrofbi

Answered 2022-10-24
Author has **12** answers

Note that ${25}^{2}=625\equiv 25mod100$, so that in fact ${25}^{n}\equiv 25mod100$ for any $n$. Because $\varphi (100)=40$, by Euler's theorem we have that ${a}^{40}\equiv 1mod100$ for any $a$ relatively prime to $$100$$ (as $$23$$ is). Thus ${23}^{41}\equiv 23mod100$. Now put these results together to find

$${23}^{41}\cdot {25}^{40}mod100.$$

$${23}^{41}\cdot {25}^{40}mod100.$$

Deborah Proctor

Answered 2022-10-25
Author has **3** answers

On applying $\phantom{\rule{thinmathspace}{0ex}}\text{}ab\phantom{\rule{thinmathspace}{0ex}}mod\phantom{\rule{thinmathspace}{0ex}}ac\phantom{\rule{thinmathspace}{0ex}}=\text{}a\phantom{\rule{thinmathspace}{0ex}}(bmodc)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}$ mod Distributive Law we get

$\phantom{\rule{thinmathspace}{0ex}}{25}^{1+J}{23}^{1+2K}mod100\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}25\phantom{\rule{thinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\underset{\equiv \text{}{1}^{J}(-1{)}^{1+2K}\equiv \text{}{3}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}{\underset{\u23df}{{25}^{J}{23}^{1+2K}}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}mod4)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}25({3})$

$\phantom{\rule{thinmathspace}{0ex}}{25}^{1+J}{23}^{1+2K}mod100\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}25\phantom{\rule{thinmathspace}{0ex}}(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\underset{\equiv \text{}{1}^{J}(-1{)}^{1+2K}\equiv \text{}{3}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}{\underset{\u23df}{{25}^{J}{23}^{1+2K}}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}mod4)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}25({3})$

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In a box there are 16 ice-creams: 6 lemon flavor,4 mint flavor and 6 strawberry flavor.When we extract two ice-creams,what's the probability of getting two different flavors,given that at least one is strawberry flavor.

That's my solution :

$P(A)$ = "Different flavors" = $\frac{{\textstyle (}\genfrac{}{}{0ex}{}{16}{2}{\textstyle )}}{16!}$

$P({B}^{c})$ = "At least one is strawberry flavor" = $1-\frac{{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )}}{16!}$

We want $P(A|{B}^{c})$ using conditional probability, where I go wrong?

That's my solution :

$P(A)$ = "Different flavors" = $\frac{{\textstyle (}\genfrac{}{}{0ex}{}{16}{2}{\textstyle )}}{16!}$

$P({B}^{c})$ = "At least one is strawberry flavor" = $1-\frac{{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )}}{16!}$

We want $P(A|{B}^{c})$ using conditional probability, where I go wrong?