MathProblemSolver101

if it's 12, then it's 12-6-3-10-5-26-13-170(which doesn't work) and a dead end there. You were correct on your intuition for 1, 2, 4, 8, and see if another pattern fits.

We know that the last application has to be $\frac{n}{2}$ because if we used $n^2 + 1 = 1,$ that means $n^2 = 0 \Rightarrow n = 0$ though that is not allowed.

So that already removes 49 numbers(remember, 1 is allowed because $1^2 + 1 =2,$ $\frac{2}{2} = 1.$

We can further note that there has to be a $2^k$ for $k \geq 0$ so then they can go like 8-4-2-1.

Intuitively, we can only have integers that are in the form $2^k$ for some number $k.$ Here's why:

Assuming we can use odd numbers, we have $2^k = k^2 + 1.$

$2 = \sqrt[k]{k^2 + 1}$

This only works with $k=1$

Then observe this graph: https://www.desmos.com/calculator/t4hcdbhc3o

You will find that the only other solution is k = 4.125, but we specified with integers only.

Furthermore, to be evenly rooted, $k^2 + 1 = i^k$ for some integer $i.$

again, the only solution is $k=1.$

Note: $1 = 2^0.$

$2 = 2^1$

etc.

This works for 2^0 -> 2^6, so your answer is $6 -0+1 = \boxed{7}$

I really liked your matching-up with constants and variables! +1

TYSM! They were all correct :D

Let $a_1$ equal Bart's madeleines, and let $a_2$ equal Jeff's madeleines.

$a_1 + 57 = a_2$

$\left(1-\frac{3}{5}\right) a_1 = \frac{1}{4} \cdot \left(1-\frac{2}{3}\right) a_2$

$\frac{2}{5} a_1 = \frac{1}{12} a_2$

$24 a_1 = 5 a_2$

$24 a_1 = 5(a_1 + 57)$

$24 a_1 = 5 a_1 + 5 \cdot 57$

$19 a_1 = 5 \cdot 57$

$a_1 = \frac{5 \cdot 19 \cdot 3}{19}$

$a_1 = 15$

$a_2 = 15 + 57 = 72$

$\frac{2}{3} a_2 = \frac{2}{3} \cdot 72 =  24 \cdot 2 = \boxed{48}$

$k - \left(\frac{1}{8} k + 14 + \frac{3}{10} \left(\frac{7}{8} k - 14 \right) + 24 + 40 \right)= 34$

$k - \frac{31k}{80} + \frac{21}{5} - 78 = 34$

$80k - 31k + 336 - 6240 = 2720$

$49k = 8624$

$k = \boxed{176}$

$n \leq 1000$

$n \equiv 7 \pmod{9}$

$n \equiv 5 \pmod{10}$

$n \equiv 9 \pmod{11}$

By the chinese remainder theorem, you have 

$n = 990k + 295$

$n \leq 1000$

$n = \boxed{295}$

You could have also noted that the units digit is simply \(\begin{align*} 3^3 &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= \boxed{7} \end{align*}\)

Too bad I can't use \equiv on begin{align*} enviorments.

Let $f(x)$ equal units digit of $x.$

$f(3^1) = 3$

$f(3^2) = 9 $

\(\begin{align*} f(3^3) &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= 7 \end{align*}\)

\(\begin{align*} f(3^4) &= 81 \pmod{10} \\ &= 1 \pmod{10} \\ &= 1 \end{align*}\)

\(\begin{align*} f(3^5) &= 243 \pmod{10} \\ &= 3 \pmod{10} \\ &= 3 \end{align*}\)

Thus, given $3^n,$

$n \equiv 0 \pmod 4, f(n) = 1$

$n \equiv 1 \pmod 4, f(n) = 3$

$n \equiv 2 \pmod 4, f(n) = 9$

$n \equiv 3 \pmod 4, f(n) = 7$

$123 \equiv 3 \pmod 4, $ so $f(n) = \boxed{7}$

I don't know if this is by all of these, but I'm assuming the a,b,c,d are not constraints for one question but multiple questions.

(a) $3 + 5 + 2 + 9 + a + 2 = 3n \Rightarrow 21 + a = 3n \Rightarrow a=0; a=3; a=6; a=9$

(b) $a+2 = 4n \Rightarrow a = 2; a = 6$

(c) All values of (b) that are also in (a), or vice versa $\Rightarrow a = 6$

(d) All values of (c) that are also in (a), or vice versa $\Rightarrow a = 6$