$+$ and $\times$ defined

In a nutshell: Ditch everything; just throw the two large numbers into a calculator and multiply them.

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Complete solution:

Throughout this solution, consider natural numbers to include $0$ (so they are $0,1,2,3,\ldots$ ).

Define $\oplus, \otimes$ to be the operations defined here, and let $+, \times$ be the addition and multiplication we know. We can prove that $\oplus$ and $\otimes$ are the usual addition and multiplication defined for natural numbers.

Observe that $S(a)$ is effectively equal to $a + 1$ , the natural number following $a$ .

We induct on $b$ to show that $a \oplus b = a + b$ . The base case $b = 0$ is done with the first axiom: $a \oplus 0 = a = a + 0$ . Now assume the induction hypothesis: $a \oplus b = a + b$ . We then have:

$a \oplus (b+1)$

$= a \oplus S(b)$ by definition of $S$

$= S(a \oplus b)$ by second addition axiom

$= (a \oplus b) + 1$ by definition of $S$

$= (a+b) + 1$ by induction hypothesis

$= a + (b+1)$ by associative property of the usual addition

Thus we have proven the inductive step: from $a \oplus b = a + b$ follows $a \oplus (b+1) = a + (b+1)$ .

We then move to multiplication. Again, we use induction that $a \otimes b = a \times b$ . The base case $b = 0$ is again done with the first axiom: $a \otimes 0 = 0 = a \times 0$ . Now assume the induction hypothesis: $a \otimes b = a \times b$ . We then have:

$a \otimes (b+1)$

$= a \otimes S(b)$ by definition of $S$

$= a \oplus (a \otimes b)$ by second multiplication axiom

$= a + (a \otimes b)$ by our result above

$= a + (a \times b)$ by induction hypothesis

$= (a \times 1) + (a \times b)$ by multiplicative identity

$= a \times (1+b)$ by distributive property of the usual multiplication over the usual addition

$= a \times (b+1)$ by commutative property of the usual addition

Thus we have proven the inductive step: from $a \otimes b = a \times b$ follows $a \otimes (b+1) = a \times (b+1)$ .

Thus we have established that the operations in the problem are none other than the usual addition and multiplication. Thus, throwing our numbers to a calculator, we obtain the result: $116180912 \times 615151219 = \boxed{71468829641331728}$ .

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Afterword: Did you notice $1,16,18,09,12$ and $6,15,15,12,19$ spell APRIL FOOLS ? That is exactly what inspiring this problem. Make a seemingly convoluted problem which turns out to be none other what we are used to. One might be interested on reading more about Peano axioms .

Let a = 116180912 y= 615151219. Let y= S(b) Thus b= 615151218. Since axS(b) = a+(a b)=a y, it give 116180912*615151219=71468829641331728.