heureka

In the prime factorization of 12! + 15! + 18! + 21! + 24!, what is the exponent of 3?

The sequence \((a_n)\) satisfies \(a_1=3\) and
\(a_{n}=\dfrac{a_{n-1}-1}{a_{n-1}+1}\)
for all \(n \ge 2\). Find \(a_{100}\)

Given that \(33^{-1} \equiv 77 \pmod{508}\),
find \(11^{-1}\pmod{508}\) as a residue modulo \(508\).
(Give an answer between 0 and 507, inclusive.)

\(\begin{array}{|rcll|} \hline 33^{-1}= \dfrac{1}{33} & \equiv& 77 \pmod{508} \\\\ \dfrac{1}{33} = \dfrac{1}{3*11} & \equiv& 77 \pmod{508} \\\\ \dfrac{1}{3*11} & \equiv& 77 \pmod{508} \quad | \quad * 3 \\\\ \dfrac{3}{3*11} & \equiv& 3*77 \pmod{508} \\\\ \dfrac{1}{11} & \equiv & 231 \pmod{508} \\\\ \dfrac{1}{11}=\mathbf{ 11^{-1} } & \equiv & \mathbf{ 231 \pmod{508} } \\ \hline \end{array}\)

Evaluate the sum \(\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} &+ \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots \\ s &=& 1*2^{-1} &+ 3*2^{-2} + 5*2^{-3} + 7*2^{-4} + 9*2^{-5} + \ldots \\ 2^{-1}s &=& &1*2^{-2} + 3*2^{-3} + 5*2^{-4} + 7*2^{-5} + \ldots \\ \hline s-2^{-1}s &=& 1*2^{-1} &+ 2 (\underbrace{2^{-2} + 2^{-3} + 2^{-4} + \ldots}_{=2^{-2}*\dfrac{1}{1-2^{-1}} } ) \\ s(1-2^{-1}) &=& 1*2^{-1} &+ 2*2^{-2}*\dfrac{1}{1-2^{-1}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*\dfrac{1}{\dfrac{1}{2}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*2 \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&*(1+2) \\ s &=& &1+2 \\ \mathbf{s} &=& \mathbf{3} \\ \hline \end{array}\)

Let a1, a2, a3, ... be an arithmetic sequence.
If a2/a4 = 3, what is a5/a3?

\(\begin{array}{|rcll|} \hline \dfrac{a_2}{a_4} &=& 3 \\ a_2 &=& 3a_4 \\ a_1+d &=& 3(a_1+3d) \\ a_1 + d &=& 3a_1+9d \\ -2a_1 &=& 8d \\ \mathbf{a_1} &=& \mathbf{-4d} \\ \hline \dfrac{a_5}{a_3} &=& x \\ a_5 &=& x*a_3 \\ a_1+4d &=& x*a_3 \quad | \quad \mathbf{a_1=-4d} \\ -4d+4d &=& x*a_3 \\ 0 &=& x*a_3 \\ x &=& \dfrac{0}{a_3} \\ \mathbf{x} &=& \mathbf{0} \\ \hline \end{array}\)

\(\dfrac{a_5}{a_3} = 0\)

Dana has only nickels in her hand,

and Ajah has exactly the same number of dimes and no other coins.

Together they have a total of 90¢. How many coins is each person holding?

1. N = D

2. N*5+D*10 = 90

    N*5+N*10 = 90

    15N = 90

    N = 6 or D = 6

If x, y, w, z are real numbers satisfying
 
\(w+x+y = -2 \\ w+x+z = 4 \\ w+y+z = 19 \\ x+y+z = 12\)

What is \(wx + yz\)?

\(\begin{array}{|lrcll|} \hline (1)& w+x+y &=& -2 \\ (2)& w+x+z &=& 4 \\ (3)& w+y+z &=& 19 \\ (4)& x+y+z &=& 12 \\ \hline (1)+(2)+(3)+(4): & 3w+3x+3y+3z &=& -2+4+19+12 \\ & 3(w+x+y+z) &=& 33 \quad | \quad : 3 \\ & \mathbf{w+x+y+z} &=& \mathbf{11} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad x+y+z=12 \\ w+12 &=& 11 \\ w &=& 11-12 \\ \mathbf{w} &=& \mathbf{-1} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+y+z=19 \\ x+19 &=& 11 \\ x &=& 11-19 \\ \mathbf{x} &=& \mathbf{-8} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+z=4 \\ y+4 &=& 11 \\ y &=& 11-4 \\ \mathbf{y} &=& \mathbf{7} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+y=-2 \\ z-2 &=& 11 \\ z &=& 11+2 \\ \mathbf{z} &=& \mathbf{13} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline wx + yz &=& (-1)(-8)+7*13 \\ &=& 8+91 \\ \mathbf{wx + yz} &=& \mathbf{99} \\ \hline \end{array}\)

Thank you, CPhill !

Find the value of \(a_2 + a_4 + \ldots + a_{98}\)

if \(a_1, a_2, a_3, \ldots\) is an arithmetic progression with common difference 1

and \(a_1 + a_2 + ... + a_{98} = 99\).

\(\begin{array}{|rcll|} \hline a_1 + a_2 + ... + a_{98} &=& 99 \\\\ \left(\dfrac{a_1 + a_{98}}{2}\right)*98 &=& 99 \\ \mathbf{a_1 + a_{98}} &=& \mathbf{\dfrac{ 2*99 }{98} } \\ \hline \end{array}\)

\(\text{Let $x=a_2 + a_4 + \ldots + a_{98}$}\\ \begin{array}{|rcll|} \hline x &=& a_2 + a_4 + \ldots + a_{98} \\\\\ x &=& \left(\dfrac{a_2 + a_{98}}{2}\right)*\dfrac{98}{2} \quad | \quad a_2 = a_1 +1 \\\\ x &=& \left(\dfrac{1 + a_1 + a_{98}}{2}\right)*\dfrac{98}{2} \\\\ x &=& (1 + a_1 + a_{98})*\dfrac{98}{4} \\\\ x &=& \dfrac{98}{4} + (a_1 + a_{98})*\dfrac{98}{4} \quad | \quad \mathbf{a_1 + a_{98}=\dfrac{ 2*99 }{98} } \\\\ x &=& \dfrac{98}{4} + \dfrac{ 2*99 }{98}*\dfrac{98}{4} \\\\ x &=& \dfrac{98}{4} + \dfrac{ 2*99 }{4} \\\\ x &=& \dfrac{98+2*99}{4} \\\\ \mathbf{x} &=& \mathbf{74} \\ \hline \end{array}\)

\(a_2 + a_4 + \ldots + a_{98} = \mathbf{74} \)

A number is a palindrome if it reads the same way forwards and backwards.
For example, 27872 and 48484 are palindromes, but 28389 and 12541 are not.
What is the sum of all of the 5-digit palindromes?

I assume:

\(\begin{array}{|l|r|r|} \hline & & \text{sum of $5$-digit palindromes} \\ \hline \text{sum until $1$-digit palindromes} & 45 & \\ \hline \text{sum until $2$-digit palindromes} & 540 & \\ \hline \text{sum until $3$-digit palindromes} & 50040 & \\ \hline \text{sum until $4$-digit palindromes} & 545040 & \\ \hline \text{sum until $5$-digit palindromes} & 50045040 & \mathbf{49500000} \\ \hline \end{array}\)