Careful! Every row is a positive integer

Call the 2-digit number in the first row $A$ , and the 2-digit number in the second row $B$ .

When these two numbers are multiplied, it must result in a multiple of $100$ .

There are $5$ different ways to factor $100$ : $100\cdot 1$ , $50\cdot 2$ , $25\cdot 4$ , $20\cdot 5$ , and $10\cdot 10$ .

This means that if $A$ was a multiple of one of those $5$ factors of $100$ and $B$ was the other factor, then when the two numbers are multiplied, the last two digits will be zeroes.

Because $A$ and $B$ must be 2-digit, $100\cdot 1$ can be ignored.

Also, because when both 2-digit numbers are multiples of $10$ , one of the 3-digit numbers in row $3$ or $4$ would end up as $0$ , $10\cdot 10$ can be ignored.

This leaves $3$ options for factoring $100$ - $50\cdot 2$ , $25\cdot 4$ , and $20\cdot 5$ .

Making $A$ and $B$ as large as possible while making sure it stays as a 2-digit number, there are $3$ options for the correct answer - $50\cdot 98$ (using $50$ and $2$ ), $75\cdot 96$ (using $25$ and $4$ ), and $80\cdot 95$ (using $20$ and $5$ ).

Out of these options, $80\cdot 95$ , or $\boxed { 7,600 }$ is the largest number.

Let us write the product of two digits numbers as $\begin{aligned} & (10a+b)(10c+d) = \text{\_\_\_}00 \\& 100(ac) + 10(ad) + 10(bc) + bd = \text{\_\_\_}00 \end{aligned}$

We can clearly be sure that to maximise the product initially the product of $a$ and $c$ should be maximum. Since $a$ and $c$ are single digit numbers so maximum value of $ac$ is either $9\times 9 =81$ or $8\times 9 = 72$ .

Case 1 : if $ac = 9\times 9 = 81$ ie $a=c$ . The product will turn out to be $\begin{aligned} & 100(9^2) + 10a(d+b) + bd = \text{\_\_\_}00 \\& 100(81) +90(b+d) + bd = \text{\_\_\_} 00\end{aligned}$ Note that there should be 2 trailling zeros and for $a=c$ then it must be $b=d=0$ . If $b=0, d \neq 0$ and vice versa there will be only 1 trailling zero. Also note that for $b=d=0$ then there exist leading zeros as $00$ in $3^{rd}$ and $4^{th}$ row so the case $ac =9\times 9$ is rejected.

Case2 : if $ac = 8\times 9 =72$ ie; $a\neq c$ . Then $\begin{aligned} 100(72) + 10(9d + 8b) + bd =\text{\_\_\_}00 \end{aligned}$ To unalter the last two zeros $bd$ must be equal zeros and $10(9d+8b) =100n, n\in\mathbb N <10$ which implies that $9d +8b = 10n$ and this equation will only be true iff $n=4, d=0$ and $b=5$ and no leading zeros case arises .

Eventually we have $a = 8 , b = 5 , c = 9$ and $d=0$ . Therefore the largest possible value of the above product is $85\times 90 = \boxed{7600}$ .

Hoping that I have done correct explanation. :)