All the numbers are here!

My method of solving this is a bit intuitive... The units digit of the remainder can be determined using the power cycle of 2 since the divisor here is 90... clearly the exponent 34^56^78 is divisible by 4 and as 2^(4K) always ends with 6, the above number also would end in 6... also, barring 12^1, every successive power of 12 is a multiple of 9 (as 12^2 = 144 is a multiple of 9)... So, the remainder also has to be a multiple of 9... So, basically, two conditions need to be satisfied here, the remainder needs to end in 6 and be divisible by 9 and of course, be less than 90... the only number satisfying all this is 36... hence, the answer...

This problem seems monstrous at first but is actually very easy to compute! Don't let the powers scare you. First, we prove the following trivial lemma.

Lemma: $\forall~x,y\in\Bbb{Z^+}\mid x\geq y\iff a^x\equiv 0\pmod{a^y}~\forall~a\in\Bbb{Z}$

Proof: $\forall~x,y\in\Bbb{Z^+}\mid x\geq y\iff \exists~ c\in\Bbb{Z_0^+}\mid a^x=a^{y+c}=a^y\cdot a^c\equiv 0\pmod{a^y}$

Hence, the lemma is proved and established.

Coming back to our problem, we compute the residue of the value in the problem modulo coprime values $2,5,9$ . We use Fermat's Little Theorem for calculating modulo $5$ and use the proved lemma in both modulo $5$ and $9$ calculations. Take the value as $X$ . Then,

$X\equiv 0\pmod{2}~~\textrm{(since 12 is even, the whole expression must be even)}$

$\large X\equiv 12^{\left(34^{56^{78}}~\bmod~4\right)}\equiv 12^{\left(2^{56^{78}}~\bmod~2^2\right)}\equiv 12^0\equiv 1\pmod5$

$12\equiv 3\pmod9\implies \large X\equiv 3^{34^{56^{78}}}\pmod{9=3^2}\implies X\equiv 0\pmod9$

These results can be summarized as follows:

$X\equiv\begin{cases}0\pmod2\\ 1\pmod5\\ 0\pmod9\end{cases}$

Using Chinese Remainder Theorem, one can easily get that $X\equiv 36\pmod{90}$ .

$\Large\therefore\quad X=12^{34^{56^{78}}}\equiv 36\pmod{90}$

Clarifications:

• The vertical bar symbol, $~"~\mid~"~$ used here denotes "such that".
• $\exists$ and $\forall$ are quantifier operators denoting "there exists" and "for all" respectively.
• $\iff$ represents the "if and only if" condition.

12^2%90=54 12^3%90=18 12^4%90=36 12^5%90=72 12^6%90=54 12^7%90=18 This pattern repeats every fourth power. Take the power to modulus 4 i.e. (34^56^78) % 4 = 0 So, the answer is (12^4%90) = 36

$90=2\cdot 3^2\cdot 5$ . Clearly $12^{34^{56^{78}}}=2\cdot 3^2\cdot k=18k$ .

By FLT, mod $5$ : $\ 12^{34^{56^{78}}}\equiv 12^{34^{56^{78}}\pmod{4}}\equiv 12^{0}$

\equiv 1\equiv 6\equiv 18k\equiv 3k\stackrel{:3}\iff k\equiv 2\,\Rightarrow\, k=5c+2 .

$12^{34^{56^{78}}}=18(5c+2)=90c+36$ .

As for Challenge Master: Find $2^{3^{4^{.^{.^{.^{9}}}}}}\bmod{10}$ .

$2^1\equiv 2,\ 2^2\equiv 4,\ 2^3\equiv 8,\ 2^4\equiv 6,\ 2^5\equiv 2,\ldots$ mod $10$ , and $3^{4^{\cdots}}\equiv (-1)^{4^{\cdots}}\equiv 1\pmod{4}$ , so $2^{3^{4^{.^{.^{.^{9}}}}}}\equiv 2^1\equiv 2\pmod{10}$ .

Working out the multiplicative cycle generated by $12\textnormal{ (mod }90)$ , we see that if the exponent is:

• $0\textnormal{ (mod }4)$ we get $36$
• $1\textnormal{ (mod }4)$ we get $72$
• $2\textnormal{ (mod }4)$ we get $54$
• $0\textnormal{ (mod }4)$ we get $18$

Similarly for the multiplicative cycle generated by $36\textnormal{ (mod }4)$ we get 0 for any exponent $\geq 2$ , so then the answer must be $\color{#20A900}{\boxed{36}}$ .

For simplicity, let $x:=\green{34}^{\blue{56}^{78}}$ . Notice that $x=4k\geq 4$ , so we have $\gcd(12^x,\:90)=18$ and cannot use Euler's Theorem directly! Instead, we factor out the common divisor $18$ , and use Euler's Theorem $(*)$ for the remaining inner modulus of $5$ : $\begin{aligned} 12^x &\equiv 18\left( \frac{12^x}{18} \mod 5 \right) \equiv 18\left( 12^{x-2}\cdot 8\mod 5 \right)\qquad \left|x=4k\qquad \left|12^4\equiv 1\mod 5\quad (*)\right.\right.\\[1.25em] &\equiv 18\left( 12^{4k-4}\cdot 12^2\cdot 8\mod 5 \right)\underset{(*)}{\equiv} 18\left( 1^{k-1}\cdot2^2\cdot 3\mod 5 \right) \equiv 18\cdot 2\equiv \boxed{36}\mod 90 \end{aligned}$

Rem.: Recall the following useful property to get rid of common divisors of base and modulus: $\begin{aligned} a&\in\mathbb{Z}, &n&\in\mathbb{N},&\gcd(a,\:n)&=g&&&\Rightarrow&&&& a\equiv g\cdot\left(\frac{a}{g}\mod \frac{n}{g}\right)\mod n \end{aligned}$

We can figure out 2 key things about 12^(34^(56^(78))):

1. Since 2m^(4n) = a number ending in 6, so must 12^(34^(56^(78))), since 34^(56^(78)) = 2a^(2b) = 4c
2. Since 3m^n (when n>1) = 9x, so must 12^(34^(56^(78)))

Thinking of 12^(34^(56^(78))) = 9m and 90x = 9n (90x being the largest multiple of 90 less than 12^(34^(56^(78)))), you can then think of the remainder as 9(m-n). Additionally (more like subtractionally, lol), the remainder is (number ending in 6) - (number ending in 0), so the remainder’s last digit is 6. Finally, a remainder is always less than the denominator: in this case, 90. Thus we need a multiple of 9 ending with 6 that is less than 90: 36

what I did was something very simple actually. 90 = 10 * 9 so now X = 12^34^56^78 mod 90 can be broken down to ((12^34^56^78) mod 10) * 9 * 9 + ((12^34^56^78 * 10 * 1)mod 9) mod 90 (using Chinese remainder theorem) k = (some other terms, but we don't care since this whole thing blows up) = 10*1

That is equal to (2^(4m) mod 10) * 9 * 9 +( (3^(34^56^78) mod 9) * k) mod 90 since 3^(34^56^78) mod 9 = 0 now it is easy to see that 2^(4m) mod 10 = 6 X = (6 * 9 * 9 mod 90) Therefore, X = (6 * 9 * 9) mod 90 = 486 mod 90 = 36 :)

First the remainder when given number is divided by 10 is 6. Since in 12^(34^(56^78)) their is 2^(4k) whose last digit i.e remainder when divided by 10is 6. Or we can say x=6mod(10). Secondly x=0mod(9). I checked for 36 x=36mod (90).As 36 leaves remainder 6divided by 10 and 0when divided by 9.Also as 10 and 9 are coprime the remainder when x divided by 90=10*9 is 36

Multiply all the exponents to get 148512 and divide it by 90. This gives a congruence of 12.

The problem becomes 12^12 Mod 90

This becomes (12^2)^6

144 Mod 90

144 is congruent to 56. 56^ 6 is congruent to 36

This becomes (36^2)^3 Mod 90

Answer is 36

$( 34^a \equiv 0 [ 4 ] ) \iff ( a = 4t)$ $while$ $t \in \mathbb{N}$ $so$ $( 34^{56^{78}} = 4t )$ $and$ $we$ $have$ $that$ $12^{4t} \equiv 36 [ 90 ]$ $so$ $12^{34^{56^{78}}} \equiv 36 [ 90 ]$