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Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.

Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters 'A', 'B', and so on to represent the digits 10, 11, and so on.

Print both the number and its square in base B.

basically we test numbers 1-300 ^2 in the base provided. (Square before converting into the base. After that then check to see if the squared number is a palindrome. If it is print out the number that is suppose to be squared in the base form and the squared number in base form if it is palindromic.

What is the program trying to do? You said something about base 11. What about it? Please tell me so I can try to fix it.

Sorry guest, please do not bring AOPS questions to this forum. That is violating the AOPS conduct.

We can solve this with patterns!

Notice how 2^2=2^1+2^0+1

2^3=2^2+2^1+2^0+1

Etc...

So this means The first k squares al added together is equal the the k+1 square - 1.

With this, the first 8 squares added together is 2^9-1 = 511, 2^10=1024, 1024-511=513

\(a+ab^2 = 40b\\ a-ab^2 = -32b\\\)

\(\text{add them}\\ 2a=8b\\ b=\dfrac{a}{4}\)

\(\text{subtract them}\\ 2ab^2 = 72b\\ \text{if }b\neq 0\\ 2ab=72\\ 2\dfrac{a^2}{4} = 72\\ a^2 = 144\\ a=\pm 12\)

\(\text{and of course }a=b=0 \text{ is also a solution}\\ \text{so }a = 0, \pm 12\)

i hope this helps

about 7.6%

\(\text{if }v=f(u) \text{ then } u = f^{-1}(v) \text{ in other words}\\ \text{if }(u,v) \text{ is a point on }f(u) \text{ then }(v,u) \text{ must be a point on }f^{-1}(v)\\ \text{thus if these two points are identical}\\ (u,v)=(v,u) \Rightarrow u=v\)

\(\text{so if }(-3,a) \text{ is on the graph of }f(x)\\ (a,-3) \text{ is on the graph of }f^{-1}(x) \\ \text{and if they coincide then }a=-3\)

\(\text{well I guess I'd simplify it all to see if anything cancels}\\ \dfrac{5}{\sqrt{80}} = \dfrac{5\sqrt{80}}{80} = \dfrac{\sqrt{5}}{4}\\ \dfrac{\sqrt{845}}{9} = \dfrac{13\sqrt{5}}{9}\\ \sqrt{45}= 3\sqrt{5}\)

so we've got

\(\sqrt{\dfrac{\frac{\sqrt{5}}{4}+\frac{13\sqrt{5}}{9}+3\sqrt{5}}{\sqrt{5}}} = \\ \sqrt{\dfrac 1 4 + \dfrac{13}{9} + 3}=\\ \sqrt{\dfrac{9+52+108}{36}} = \sqrt{\dfrac{169}{36}}= \dfrac{13}{6}\)

\(\text{the inscribed sphere has a radius }\tilde{r} = 1\\ vol = \dfrac 4 3 \pi r^3 = \dfrac 4 3 \pi\)

\(\text{axis of symmetry at }x=-3 \Rightarrow\\ a(x+3)^2 + c = ax^2 + 6ax+ (9+c)\\ \dfrac{b}{a} = \dfrac{6a}{a} = 6\)

The cheater way to do it is to remember that the axis of symmetry is at

\(x = -\dfrac{b}{2a} \\ \text{so here}\\ -3 = -\dfrac{b}{2a} \\ \dfrac{b}{a} = (-3)(-2) = 6\)

\(p = \dfrac{|M\cap D|}{|M|}\\ M = \{1,5\}\\ D = \{1, 2, 5\}\\ M\cap D = \{1, 5\}\\ p = \dfrac{|\{1,5\}|}{|\{1,5\}|} = \dfrac{2}{2}=1\)

\(\text{The odds are infinite}\)

\(\text{# of heads out of 10 is binomially distributed }n=10, ~p=\dfrac 1 2\\ A \sim \{\text{7 heads}\}\\ B \sim \{\text{at least 3 heads}\}\\ P[A|B] = \dfrac{P[B|A]P[A]}{P[B]} \)

\(P[B|A] = 1\\ P[A]=\dbinom{10}{7} 2^{-10} = \dfrac{15}{128}\\ P[B] = 1-\left(\sum \limits_{k=0}^2~\dbinom{10}{k}2^{-10}\right) = \dfrac{121}{128}\)

\(P[A|B] = \dfrac{1 \cdot \frac{15}{128}}{\frac{121}{128}} = \dfrac{15}{121}\)

\(\text{odds} = \dfrac{P[A|B]}{1-P[A|B]} = \dfrac{15}{121-15} = \dfrac{15}{106}\)

\(\text{The point will have it's }x \text{ coordinate at the midpoint of the given points}\\ x=4\\ \text{the distance from the point to say }(0,5) \text{ is equal to the (squared) distance between the given points}\\ (4-0)^2+(y-5)^2 = 8^2\\ 16 + (y-5)^2 = 64\\ (y-5)^2 = 48\\ y = \pm \sqrt{48}+5 = \pm 4\sqrt{3}+5\\ \text{the point that lies in the first quadrant is thus }\\ (x,y) = (4, 4\sqrt{3}+5)\)

I don't know what programming language you are learning. But, for this question, you would set up a simple search routine that would subtract A^2 or B^2 from the given number on the RHS and take its square root. If it returns an integer, then that is one number found. And so on....

In C++ the routine would look something like this:

n=1; a=(2#(1863702780625 - n^2);b=if(ceil(a) - floor(a)==0, goto3, goto5);c=(2#(1863702780625 - n^2);print"n=",n, "=",c;   n++;if(n<=1000000, goto1, discard=0;

With the above routine, my computer listed these numbers:

1 - 33292^2 + 1364769^2
2 -126945^2 + 1359260^2
3 -160055^2 + 1355760^2
4 -193060^2 + 1351455^2
5 -225960^2 + 1346345^2
6 -295008^2 + 1332919^2
7 -350175^2 + 1319500^2
8 -359639^2 + 1316952^2 
9 -382249^2 + 1310568^2
10 -502460^2 + 1269345^2
11 -511560^2 + 1265705^2
12 -563745^2 + 1243340^2
13 -627039^2 + 1212652^2
14 -656425^2 + 1197000^2
15 -685412^2 + 1180641^2
16 -705628^2 + 1168671^2
17 -714000^2 + 1163575^2
18 -819105^2 + 1092140^2
19 -845495^2 + 1071840^2
20 -917112^2 + 1011241^2
21 -941500^2 + 988575^2
22 -965321^2 + 965328^2

Note: You have to be wary of duplicates!

What the formula for the area of a square?

Area =Side^2

Area =(2sqrt(3))^2

Area =4 x 3 =12 units^2

Thank you

That is correct in base 10: You can list all the numbers in base 6 as follows:

(1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, 30, 31, 32, 33, 34, 35, 40, 41, 42, 43, 44, 45)=29 numbers.

Then sum them in base 10 as shown above =435 in base 10. 

Convert 435 from base 10 to base 6:

435 / 6 = 3 as the remainder.

72 / 6   = 0 as the remainder.

12 / 6   = 0 as the remainder.

2/6       = 2 as the remainder.

so 435 in base 10 =2003(from the bottom up) in base 6, which is the answer to your question.

Note: Not many people here feel competent enough to answer questions in Statistics. I entered some of your numbers in WolframAlpha and got some results on this page. Look at the bottom of the page under "percentile" and see if it gives meaningful answers to your question.

https://www.wolframalpha.com/input/?i=mean%3D7.2,+standard+deviation+%3D2.1,++95%25+confidence+interval

There are 23 such numbers as follows:

(100, 121, 144, 169, 196, 200, 225, 242, 256, 288, 289, 300, 324, 338, 361, 363, 392, 400, 432, 441, 450, 484, 500) =23 such numbers.

You can write each number as: 100 =10sqrt(1), 200 =10Sqrt(2), 288=12sqrt(2)......and so on.

By trial and error, I found 17, 27, and 67. 

17 x 27 x 67 = 30,753                      This satisfies the "tens digit of the product is 5" requirement

17 + 27 + 67 = 111                           The sum of the three numbers

The tens digit of the sum is 1  

Remember I said I found these by trial and error.  There might be other solutions. 

Oh ok thank you soo much!

See the answer here by Max Wong:

https://web2.0calc.com/questions/the-sum-of-a-certain-infinite-geometric-series-is#r4