+169 We know the last two digits can only result in a number lower than 100, we know multiplication between two large numbers is basically an accumalation of simple multiplications. In this case we have (1000+100+20+2)*(1000+100+20+2), since we can only allow numbers smaller than 100 we can drop 1000 and 100. We get (20+2)*(20+2) which results in 400+40+40+4, 400 is greater than 100, so 40+40+4=84. 84 are the last two digits. Hope this helps
+118677 I only just realised that Honga had answered this. thanks Honga.
Alan answered a similar question only a couple of days ago. This is the thread.
http://web2.0calc.com/questions/443-443-what-are-the-last-three-digits
+129852 But we're not taking (1122) * (1122). We're taking 1122 and raising to the power of 1122 factorial....!!! (If I read the question correctly...)
I don't know how to determine this !!! Whatever this number is, it is HUGE!!!
+169 D**n, multiplication seemed logical to me, nevertheless this should be able to be figured out as well.
The last number is relatively easy since it's the result of $${{\mathtt{2}}}^{{\mathtt{n}}}$$, which means the last digit regularly skips between this array of numbers: 2,4,8,6.
For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}\right)}$$ the last digit is 6.
For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}$$ the last digit is 2.
For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ the last digit is 4.
For $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}$$ the last digit is 8.
We know the last digit is multiplied 1122 times, $${\frac{{\mathtt{1\,122}}}{{\mathtt{4}}}} = {\frac{{\mathtt{561}}}{{\mathtt{2}}}} = {\mathtt{280.5}}$$ so it is of the form $${{\mathtt{2}}}^{\left({\mathtt{4}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$ which means the last digit is 4.
While doing this I figured out a better way to solve this problem, we know this of the exponent:
$${\mathtt{1\,122}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{17}}$$
Another thing we know for sure is that while multiplying two numbers only the last two digits of each number affect the last two digits of the answer. Hence all digits save for the last two are irrelevant for further multiplication.
Since only the last two digits matter of the 1122 being multiplied we can drop the first two digits:
$${{\mathtt{22}}}^{{\mathtt{1\,122}}} = \left({\left({\left({\left({{\mathtt{22}}}^{{\mathtt{2}}}\right)}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}\right)$$
$${\left({\left({{\mathtt{484}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$
Only last two digits matter
$${\left({\left({{\mathtt{84}}}^{{\mathtt{3}}}\right)}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$
$${\left({{\mathtt{592\,704}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$
Only last two digits matter
$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}}$$
$${\left({{\mathtt{4}}}^{{\mathtt{11}}}\right)}^{{\mathtt{17}}} = {\left({\mathtt{4\,194\,304}}\right)}^{{\mathtt{17}}}$$
Only last two digits matter
$${{\mathtt{4}}}^{{\mathtt{17}}}$$
$${{\mathtt{4}}}^{{\mathtt{17}}} = {{\mathtt{4}}}^{{\mathtt{11}}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {\mathtt{4\,194\,304}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}}$$
Only last two digits matter
$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{6}}} = {{\mathtt{4}}}^{{\mathtt{7}}} = {\mathtt{16\,384}}$$
Last two digits of $${{\mathtt{1\,122}}}^{{\mathtt{1\,122}}}$$ are 84, case closed.
+118677 WOW Honga,
This is really impressive.
To be truthful I spend so much time on administrating this site that i do not always look at answers that I am sure I would find greatly illuminating.
I am very glad that I looked at this answer. Your explanation is brilliant.
I hope that you stick around to WOW us a lot more.
Thank you.
