A Detail in one of Led Zeppelin's Albums

$\begin{array}{lllll} & & & ? & ? & ? & ? \\ & & \times & & ? & ? & ? \\ \hline & & 3 & 3 & 2 & 4 & 0 \\ & 7 & 7 & 2 & 4 & 0 & + \\ 2 & 5 & 0 & 8 & 0 & + & + \\ \hline ? & ? & ? & ? & ? & ? & ? \end{array}$

The multiplication above is the same multiplication from the problem except that we have plus signs instead of zeroes. These zeroes are optional.

Each term of this sum is a multiple of the first factor of the multiplication. So, the Greatest Common Divisor of these must give us the first factor or a multiple of this factor.

$\text{gcd} (32240,77240,25080)=40$

The gcd of these three numbers is 40. As the first factor of the multiplication must be a 4-digit number, we can't have these terms in the final sum. So the entire multiplication is wrong.

Method 1: Suppose this is possible. Then these 3 intermediate products can be expressed as $\begin{cases} a\times b = 33240 \\ a \times c = 77240 \\ a\times d = 25080, \end{cases}$ where $a,b,c,d$ are positive integers and $b,c,d$ are single digits.

This also means that if we take the ratio of any of the equations, the following equations must be able to be expressed as the ratio of single digit integers. $\dfrac{a\times b}{a\times c} = \dfrac{33240}{77240} , \qquad \dfrac{a\times b}{a\times d} = \dfrac{33240}{25080} , \qquad \dfrac{a\times c}{a\times d} = \dfrac{77240}{25080} .$ However, upon simplifying these fractions, we see that not all of these 3 fractions can be expressed as the ratio of single digit integers. A contradiction! Hence, this is not possible.

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Method 2: The summand is equal to $33240 + 772400 + 2508000=3313640$ . Now, suppose a solution exists, and since there are 3 intermediate products, then this summand must have a 3-digit divisor. Prime factorization shows that $3313640 = 2^3\times5\times11\times17\times443,$ so the only possible 3-digit factor could be $443$ or $443\times2=886$ .

Let $A$ and $B$ be the two integers being multiplied together such that their product is 3313640.

If one of $A$ and $B$ is 443, then the other must be $3313640\div443=7480$ , however, by expanding the long multiplication, we are not able to get the 3 intermediate products.

Likewise, if one of $A$ and $B$ is 886, then the other must be $3313640\div886=3740$ , however, by expanding the long multiplication, we are not able to get the 3 intermediate products either. A contradiction!

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Method 3: Suppose this is possible. Then these 3 intermediate products can be expressed as $\begin{cases} a\times b = 33240 \\ a \times c = 77240 \\ a\times d = 25080, \end{cases}$ where $a,b,c,d$ are positive integers and $b,c,d$ are single digits.

Taking the difference of 2 of these equations give $a\times b - a \times d = a(b-d) = 77240 - 33240 = 44000$ . Since $b$ and $d$ are single digits, so is $b-d$ . 44000 factors to $44000 = 2^5\times5^3\times11$ , thus the only possible values of $b-d$ are $2,4$ or 5 only. So the only possible values of $a$ are $22000, 11000$ or $8800$ only. But checking back shows that $b,c,d$ cannot be all integers for any of these $a$ 's, which is absurd.

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Method 4: Suppose this is possible. Then these 3 intermediate products can be expressed as $\begin{cases} a\times b = 33240 \\ a \times c = 77240 \\ a\times d = 25080, \end{cases}$ where $a,b,c,d$ are positive integers and $b,c,d$ are single digits.

By divisibility rules of 3 , we can see that among these 3 intermediate products, only 77240 is not divisible by 3. Thus, neither $a$ nor $c$ is divisible by 3. However, the other 2 intermediate products are divisible by 3, so both $b$ and $d$ are divisible by 3.

Also note that $a\times d < a\times b < a\times c$ , so $d <b<c$ must be fulfilled.

Combining the information we gathered from the 2 paragraphs above, the only possible solutions of $(d,b,c)$ are $(3,6,7), (3,6,8)$ . Thus, 3-digit factor of the summand $33240 + 772400 + 2508000=3313640$ could be either $\overline{dcb} = 100d + 10c + b = 376$ or $386$ . However, neither 376 nor 386 is a factor of the summand 3313640. A contradiction!