Equal Products

We can show easily that 0, 5 and 7 can't take there.

So we get $A \times B \times C \times D \times E \times F \times G$ = $1 \times 2 \times 3 \times 4 \times 6 \times 8 \times 9$ = $1 \times 2 \times 3 \times 2^{2} \times 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 3$ = $2^{7} \times 3^{4}$

Since $A \times B \times C$ = $D \times E \times F$

=> $(A \times B \times C)^{2} \times G$ = $(2^{3} \times 3^{2})^{2} \times 2$ or $(2^{2} \times 3^{2})^{2} \times 8$

So, G = only either 2 or 8.

But we can't take G = 8 as then $A \times B \times C$ = 36

And we know $A \times B \times C$ = $B \times G \times E$

=> $B \times E$ = $\frac{ A \times B \times C}{G}$ = $\frac{36}{8}$ = 4.5 that is not an integer, so not acceptable.

So, G = 2 is the answer.

Answer:

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If any of these values equals to 0, then $A\times B \times C = B\times G \times E = D \times E \times F = 0$ .

But since only one of them can be 0, then at least one of $A,B,C$ is equals to 0 and at least one of $D,E,F$ is equals to 0. But this is a contradiction because exactly one of them can be equals to 0 only.

Similarly, iIf any of these values equals to 5, then $A\times B \times C , B\times G \times E ,D \times E \times F$ ends with a value of 5 or 0.

But since only one of them must be divisible by 5, then at least one of $A,B,C$ is 5 and at least one of $D,E,F$ is equals to 5. But this is a contradiction because exactly one of them can be equals to 5 only.

Analogously, we can show that none of them can be 7.

We can finish the entire puzzle off quickly after knowing that 0,5,7 can't be the possible values.

0 is obviously not the answer.I then eliminated 3,9 and 4 using contradiction to get the answer.

As other solutions have described, there are a number of ways to prove that none of the digits can be 0, 5, or 7. I found the simplest way to conceptualize this is to be:

• As 7 is prime and has no single-digit multiples, it is impossible for these expressions to be equal unless they all include 7 or none of them include 7.
• The same is true for 5.
• Similarly for 0, it is impossible for these expressions to be equal unless they all include 0 or none of them include 0.
• Since there is no common integer shared by all 3 expressions, we can rule out any of the integers being 0, 5, or 7.
• Therefore, the 7 integers used are 1, 2, 3, 4, 6, 8, and 9.

Then let n = $A \times B \times C$ = $B \times G \times E$ = $D \times E \times F$

We can infer that n must be factorable by all seven of these integers. The lowest common multiple for these numbers is 72. Conveniently, there are three possible ways to factor 72 into three distinct integers are:

• $9 \times 8 \times 1$
• $9 \times 4 \times 2$
• $6 \times 4 \times 3$

Looking at the initial equation, we can see that B and E are used twice. Since 9 and 4 are the only integers that appear twice in the above list, we can infer that B = 9 or 4, and E equals 9 or 4 (whichever is not B). Thus:

• $n = B \times G \times E$
• $72 = 9 \times G \times 4$
• $\boxed{2 = G}$

Note that this solution does not prove that n must equal 72; only that there is a legal solution in which n = 72 and G = 2.

This was trickier than I thought it was at first glance. I first eliminated the obvious 0 as Dinesh did. And as Dinesh did, I saw that 5 and 7 can't go anywhere because there weren't multiple 5s and 7s to support the other multiplications.
I spent longer than I thought I would on what the product would be. Perhaps I convinced myself it was 48 and couldn't make anything work. When I finally broke out the remaining numbers {1,2,3,4,6,8,9} to the product of primes, it was clear that I needed to 2^3 and 3^2 in all of them.

ABC = BGE => AC = GE BGE = DEF => BG = DF

all are distinct numbers. possible combinations of AC ,GE and BG, DF will be

3 * 2 , 6 * 1 and

4 * 2 , 8 * 1 and

4 * 3 , 6 * 2 and

3 * 6 , 9 * 2 and

4 * 6 , 8 * 3

common integer in the ablove 2 combinations is G,

possible combinations of AC ,GE and BG DF are

3 * 2, 6 * 1 and 4 * 2, 8 * 1 => 2 common numbers (1,2)

3 * 2, 6*1 and 4 * 3, 6 * 2 => 2 common numbers (3,2)

3 * 2, 6 * 1 and 6 * 3, 9 * 2 => 3 common numbers (3,2,6)

3 * 2, 6 * 1 and 4 * 6, 8 * 3 => 3 common numbers (3,2'6)

4 * 2, 8 * 1 and 4 * 3, 6 * 2 => 2 common numbers (4,2)

4 * 2, 8 * 1 and 6 * 3, 9 * 2 => 1 common number (2)

4 * 2, 8 * 1 and 4 * 6, 8 * 3 => 2 common numbers (4,8)

4 * 3, 6 * 2 and 3 * 6, 9 * 2 => 2 common numbers (6,2)

4 * 3, 6 * 2 and 4 * 6, 8 * 3 => 3 common numbers (4,6,3)

Answer must be 4 * 2, 8 * 1 and 6 * 3, 9 * 2 => which have only one common number (2) which is G so the answer is obviously 2.

A x B x C = B x G x E > A x C = G x E B x G x E = D x E x F > B x G = D x F

I found the answers by trial A=4 B=5 C=3 D=1 E=6 F=10 G=2

4 x 3 = 2 x 6 5 x 2 = 1 x 10

we should look out for 7 numbers because thaere are only 7 alphabets. Find the only number which is divisible by seven DISTINCT numbers. By calculation the oddsv we get 72 as that number. 72 has- 1,2,3,4,6,8 and 9 divisible to it. then you rearrange these numbers to get 3 1 4 2 9 6 8

• 1) None of these digits can be 0, because one product will turn to zero while others won't
• 2) Of all digits exclude also 2 highest primes: 5 and 7. Because again if one product contains any of these, then others won't be able to have them.
• 3) We have only 1,2,3,4,6,8,9 left and all these numbers can be represented as products of 2 and 3.
• 4) Consider numbers 3,6,9 as they are the only source of 3s. B=9 D=3 F=6 (or any simmetrical allocation) is the only option which will provide each of the products with exactly 3*3.
• 5) Now it is easy to allocate all the remaining numbers with G=2 being the only option.

You just take A B C and consider them a single digit like 2 and do the same to D E F after that you get this 2x2x2=2xGx2=2x2x2 Is approximately 8=4xG=8 and if we consider G to be 2 then the equation is right but it only works if you use 2 if you use 3 then you will get 3 i know a bummer right?

This may help: https://www.desmos.com/calculator/2jls3fklcs

Here's how I solved the puzzle: The puzzle takes 7 digits, so three must be eliminated. Eliminate digits that don't contain any common factors (besides 1) with other digits since there will be no way to replicate that product in another row or column. That means 0, 5 and 7 are out. Now if you look for digits with common factors, let a = 2 and b = 3. Then 4 = a^2, 6 = a b, 8 = a^3 and 9 = b^2. To solve the puzzle, each row and column needs the least common multiple (LCM) of these numbers which is (a^3) (b^2) = 72.

You can view my solution here: https://www.desmos.com/calculator/mtglzsqk8j

Let's denote A x B x C = B x G x E = D x E x F = P.

It must be possible to "divise" P into at least 5 integers:

P = a * b * c * d * e

in order that these 5 can be "re-grouped" into 3 in at least two mutually exclusive different ways:

P = (a * b) * (c * d) * e = a * (b * c) * (d * e) = A x B x C = D x E x F

otherwise, e.g. if for example it is only possible to have

P = a * b * c * d

then it cannot be the product of two mutually exclusive sets of 3 integers (ABC and DEF).

Therefore, at least two of each of {A, B, C} and {D, E, F} have to be composite (non-prime) integers.

There are only four composite integers available: 4, 6, 8, 9.

If we group this into {4, 6} and {8, 9}, and add to these sets respectively 3 and 1:

3 x 4 x 6 = 1 x 8 x 9 = 72

From the remaining integers: {2, 5, 7}, 72 is divisible only into 2, therefore:

3 x 4 x 6 = 4 x 2 x 9 = 1 x 9 x 8

G = 2