The Number Following 10 and preceeding 12

If you are unfamiliar with modular arithmetic or congruence , then:

$a \equiv b \pmod{n}$ is read as " $a$ is congruent to $b$ modulo $n.$ " It means that $(a-b)$ is divisible by $n.$

$\begin{aligned} 21 &\equiv -1 \pmod{11} \\ 12 &\equiv 1 \pmod{11} \end{aligned}$

You can raise each side of a congruence to the same exponent, and the congruence still holds .

$\begin{aligned} 21^{100} &\equiv (-1)^{100} \pmod{11} \\ 21^{100} &\equiv 1 \pmod{11} \\ \\ 12^{100} &\equiv 1^{100} \pmod{11} \\ 12^{100} &\equiv 1 \pmod{11} \end{aligned}$

We can substitute just like with equations:

$\begin{aligned} 21^{100} -12^{100} &\equiv 1-1 \pmod{11} \\ 21^{100} -12^{100} &\equiv 0 \pmod{11} \\ \end{aligned}$

Using the definition of congruence, we can interpret this as $21^{100} -12^{100}$ is divisible by 11. In fact, as long as the exponent on 21 is even, the difference will always be divisible by 11. For example, $21^6-12^7$ is divisible by 11.

Just for fun, here's a solution that relies on the fact that $a - b|a^k - b^k$ :

$\begin{aligned} 21^{100} - 12^{100} &= (21^2)^{50} - (12^2)^{50} \\ &= (21^2 - 12^2)(d) \\ &= (21 + 12)(21 - 12)(d) \\ &= 11(3)(9)(d), \end{aligned}$

where $d$ is some expression that evaluates to a positive integer. We conclude that yes , $21^{100} - 12^{100}$ is divisible by 11.

Using binomial expansion, $21^{100} = (22 - 1)^{100} = 22^{100} + 100\cdot 22^{99}\cdot (-1)^1 + \cdots + \binom{100}{k} 22^k (-1)^{100-k} + \cdots + 100\cdot 22^1\cdot (-1)^{99} + (-1)^{100},$ and $12^{100} = (11 + 1)^{100} = 11^{100} + 100\cdot 11^{99}\cdot (+1)^1 + \cdots + \binom{100}{k} 11^k (+1)^{100-k} + \cdots + 100\cdot 11^1\cdot (+1)^{99} + (+1)^{100}.$ Each term in these expansions is a multiple of 11, except for the last term of each expression, which is 1.

Since we subtract two numbers that are one more than a multiple of 11, $21^{100} - 12^{100}$ is a multiple of 11.

In this solution I have used some basic properties of Modular Arithematic.

$(21^{100}-12^{100})\mod(11)\equiv[21^{100}\mod(11)]-[12^{100}\mod(11)]$

$21^{100}\equiv(22-1)^{100}\equiv(-1)^{100}\equiv\boxed{1}\mod(11)$

$12^{100}\equiv(11+1)^{100}\equiv(1)^{100}\equiv\boxed{1}\mod(11)$

Therefore $(21^{100}-12^{100})\mod(11)\equiv[21^{100}\mod(11)]-[12^{100}\mod(11)]\equiv(1-1)\mod(11)\equiv\boxed{0}\mod{11}$

So the answer is 'Yes'.

$21^{100}-12^{100}=3^{100}(7^{100}-4^{100})=3^{100}(7^{50}-4^{50})(7^{50}+4^{50})=3^{100}(7^{50}-4^{50})(7+4)(7^{49}-7^{48}4^{1}+7^{47}4^{2}-...+4^{49})$ $7+4=11$ , so the given number is divisible by 11

Slightly different from the other solutions I see here, but of the same ilk.

Straight factoring gives us: $x^{100}-y^{100}=(x^{50}-y^{50})(x^{50}+y^{50})$

Likewise, $x^{50}-y^{50}=(x^{25}-y^{25})(x^{25}+y^{25})$

Then, $x^{25}+y^{25}=(x^{5}+y^{5})(x^{20}-x^{15}y^{5}+x^{10}y^{10}-x^{5}y^{15}+y^{20})$

Finally,

$x^{5}+y^{5}=(x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})$

In conclusion, $(x+y) | (x^{100}-y^{100})$ .

Specifically, $11|33|(21^{100}-12^{100})$ .

$21^{100} - 12^{100} = (11+10)^{100} - (11+1)^{100}$ $\\$ $\text{If we expand the R.H.S, only term that doesn't inlude '11' is} "10^{100} - 1^{100}"$ $\\$ $10^{2n} -1 \text{ is divisible by 11 } \forall \text{ n } > 1$

I may be wrong, but here goes: 21 x 12 = 252 and 252 is divisible by 12 therefore raising 21 and 12 by power 100 and subtracting the results would also be divisible by 12. That is why I marked the YES option, and lo and behold it was correct, it surprised me so I have shared my take,but please don't make fun.

(A^n-B^n) is always divisible by A-B and divisible by A+B if n is even. So the given expression is divisible by 21+12=33, which is divisible by 11. Keep solving :)

$21^{100} - 12^{100} = 3^{100}(7^{100} - 4^{100})$

$3^{100}(7^{100} - 4^{100}) = 3^{100}(7^{50} + 4^{50})(7^{50} - 4^{50})$

$3^{100}(7^{50} + 4^{50})(7^{50} - 4^{50}) = 3^{100} \times 11^{50} \times 3^{50}$

$3^{100} \times 11^{50} \times 3^{50} = 3^{150} \times 11^{50}$

So yes, the result is divisible by 110.

Since a^n - b^n is divisible by (a+b) when n is even, (21^100 - 12^100) is divisible by 21 + 12 = 33, and thus by 11.

Challenge: Is 5555^2222 + 2222^5555 divisible by 7?

3^100(7^100-4^100)=21^100-12^100. For every other integer greater than zero, 7^n-4^n is divisible by 11. 100mod(2)=0, so (7^100-4^100) is divisible by 11, and 21^100-12^100 is divisible by 11.