Another unusual cryptogram

Notice that,

Firstly, to fill in the boxes with the digits 2, 3, 4, 5, 6, 7 with no digit repeated, in general:

1. The first box and the third box can never be 5, because it will result in a number ending with either 0 (which is not an option) or 5 (which would be a repeat of the digit 5). This leaves us with 2, 3, 4, 6, and 7 as the possible digit for the first box.

2. Consequently, the sixth box can never be 5 as well, because only numbers ending with 5 will produce a number wih 5 as the last digit.

3. The sixth box can never be 3, because none of the products between any two of 2, 4, 5, 6, and 7 will produce a number with 3 as the last digit.

4. The sixth box can never be 7, because only multipliers ending with 1 and 7 will produce a number with 7 as the last digit. However, 1 is not an option.

5. Since the lowest digit is 2, the products of the two integers must be greater than 234, which is the least possible number to make up from the 6 available digits.

While keeping in mind the general requirements that I have analysed above, we shall now proceed on to the case by case scenarios:

• When the first box is 2: there is no need to waste your time checking all of the combinations, since the largest possible product is 2 x 76 = 152, which do not satisfy general requirement #5.

• When the first box is 3: similarly, there is no need to waste your time either. The largest possible product is 3 x 76 = 228, which do not satisfy general requirement #5.

• When the first box is 4:

  1. The third box can only be 3, because 4 x 2 and 4 x 7 will produce an 8. Meanwhile, 4 x 4 is impossible because there is a repeat of 4 and 4 x 6 will result in a last digit of 4, which will be a repeat of 4 too.

  2. Consequently, the sixth box can only be 2 as a result of the only possible combination of 4 x 3 mod 10 = 2.

  3. Hence, these leave us with the options of 5, 6, and 7 for the second box since the digits of 2 and 3 are already used in other boxes.

     While keeping all the above-mentioned things in mind, we shall now check the possible combinations of when the first box is 4:

     4 x 53 = 212

     4 x 63 = 252

     4 x 73 = 292

     All of which failed to satisfy the basic requirement given by the question, since there is a repeat of the digit 2.

• When the first box is 6:

  1. The third box can only be 7, because 6 x 2 and 6 x 4 will result in a repeat of the last digit 2 and 4 respectively, and 6 x 3 will result in a last digit of 8, which is not an option.

  2. Consequently, the sixth box can only be 2 as a result of the only possible combination of 6 x 7 mod 10 = 2.

  3. Hence, this leaves us with the options of: 3, 4, and 5 for the second box since the digits of 2 and 7 are already used in other boxes.

     While keeping all the above-mentioned things in mind, we shall now check the possible combinations of when the first box is 6:

     6 x 37 = 222 (repeated 2)

     6 x 47 = 282 (repeated 2)

     6 x 57 = 342 (satisfies all the given and derived conditions)

Yay! A solution! Now, we will need to find the other solution, which we will find when the first box is 7.

• When the first box is 7:

  1. The third box can only be 2 or 6, because 7 x 3 and 7 x 4 will result in a last digit of 1 and 8 respectively, which are not in the available options.

  2. Consequently, the sixth box can only be 4 or 2 as a result of 7 x 2 mod 10 and 7 x 6 mod 10 respectively.

  3. Hence, these leave us with the options of 3, 5, and 6 when the third box is 2, and the options of 3, 4, and 5 when the third box is 6.

     While keeping all the above-mentioned things in mind, we shall now check the possible combinations of when the first box is 7:

     7 x 32 = 224 (repeated 2)

     7 x 52 = 364 (satisfies all the given and derived conditions)

     ... though we have already found the solution, there is no hurt checking the other possible combinations, right? Those are:

     7 x 62 = 434 (repeated 4)

     7 x 36 = 252 (repeated 2)

     7 x 46 = 322 (repeated 2)

     7 x 56 = 392 (impossible 9)

Therefore, we can conclude that the only 2 possible solutions for this problem are 6 x 57 = 342 and 7 x 52 = 364

Adding both of the three-digit integers, we have 342 + 364 = 706 .

The two solutions are $6 \times 57 = 342$ and $7 \times 52 = 364$ .

Regarding how to find these solutions, I did it with mostly trial and error, so it's your chance to post a solution if you have a cleaner way.

possible units place numbers are:

• 6 & 7, (2)
• 7 &2, (4)
• 3 &4 (2)
• 3 & 2 (6) Then you fill in the digits you know,

7 * a 6 = x y 2

brute force to check if any of the other numbers would fit in, if it doesn't work discard the possibility and try the next one.

6 * a 7 = x y 2 'a' cannot be 7 or 6 as they're taken. The next number is 5 so:

6 * 5 7 = 3 4 2 and voila!! first solution

You need to work methodically!

Next iteration of brute force:

7 * a 2 = x y 4

'a' cannot be 7 as it is taken, 6 next:

7 * 6 2 = 4 3 4, which is not possible so try 5:

7 * 5 2 = 3 6 4

There we go!!

3 4 6 + 3 4 2 = 7 0 6

And that is your answer!!

There are 4 possibilities for the numbers' last digits.
1) 2 x 3 = #6.
2) 2 x 7 = #4.
3) 3 x 4 = #2.
4) 6 x 7 = #2.

We see that 5 is nowhere on the list, so it's either in the tens or hundreds places. Let's check the upper limit for the multiplication results. The most of the possible products should be 7 x 62 = 434 ≤ 500, so 5 cannot be the hundreds of the 3 digits number either. Now it's just down to the tens places for 5. Since 2 will be used as a tail of the numbers in any case, so it can't also be in the hundreds place. We are left with only 3 or 4 to take their place there, so the third possibility of 3 x 4 = #2 is now void and nullified.
Looking back at the most possible product we got earlier, it started with 4 at hundreds and 3 at tens. This looks like the end of our 400+ product quest, since the lesser and least digit of 2 can't be used as substitute there and 434 itself is invalid by the repeated digits of 4, too. Our next best attempt at maximal product had to be 7 x 56 = 392, already less than 400. Thus, we had it confirmed that 3 must be at the hundreds place, and by this fact itself we can further invalidate the first possibility.

1) 2 x 3 = #6 ==> invalid
2) 2 x 7 = #4
3) 3 x 4 = #2 ==> invalid
4) 6 x 7 = #2

Now, separate the digits already taken (those that came last in the three numbers even though 2 of them are still interchangeable) and yet to be used ones.

1) 2 x 3 = #6 ==> invalid
2) 2 x 7 = #4 { Waiting list : 5 & 6 , Taken list : 3 for hundreds}
3) 3 x 4 = #2 ==> invalid
4) 6 x 7 = #2 { Waiting list : 4 & 5 , Taken list : 3 for hundreds}

Note that 2 x 7 = 10 + 4 and as such would send a carryover of an odd value while 6 x 7 = 40 + 2 and as such would send a carryover of an even value. As both of the multiplications involved carry-overs from the ones digits onto the tens of 1 and 4 in value, we can exclude some combos like 40 (choosing between 4 & 5) x 6 + 40 (carryover from possibility 4)) = 280 < 300 or 50 (choosing between 4 & 5) x 7 + 40 (carryover from possibility 4)) = 390 > 340 (4 from the leftover of the earlier chosen 5), although for possibility 2) the fact that 7 has to be the single digit number is quite obvious (or the resulting 3 digit number won't even reach 200).

Now we try 7 x 52 , 6 x 57 , 7 x 46 & 7 x 42 for starters.
1) 7 x 52 = 364 [✓]
2) 6 x 57 = 342 [✓]
3) 7 x 46 = 322 [X]
4) 7 x 42 = 294 [X]

As we have already reached the lower limit of < 300, then we can safely stop now and proceed with the final answer of
= 7 x 52 + 6 x 57
= 364 + 342
= 706

I know the problem is old and solved, but my method was simple and involved very little brute force, so I thought it worth sharing.

The crux of my method is the possible triples of ones digits. Using these six numbers, there are only 4 distinct possibilities:

$2*3=6$

$3*4=2$

$6*7=2$

$2*7=4$

The other digit we want to examine is #4, the hundreds digit. The largest number possible is $7*65=455$ , so 5, 6, and 7 are out. Also, 2 is present in all of the ones combinations, so it cannot be the hundreds digit.

Finally, assume the hundreds digit is 4. This eliminates two of the ones combos, and the largest numbers possible for the remaining sets are 342 for 6-7-2*, and 146 for 2-3-6. Neither is over 400, so the fourth digit cannot be 4. Both of our answers are in the 300s.

*This is actually one of the solutions, but I only realised that I found it so early while writing this solution. I only found the other at the end, so keep reading.:)

Now, at this point, there aren’t many combos left. Two of the ones triples have been eliminated for containing 3, so you only have two triples left. These give four possibilities, one of which can be removed since 2 can’t be the first digit. Each of these 3 has two subposdibilities, for a total of six which can all be checked in about 30 seconds with a calculator. Doing this, you’ll find that the two possibilities are $7*52=364$ and $6*57=342$ . Adding, we get the answer of 706.

My solution is as follows : firstly in order to simplify a brute force search of all the numbers find the values for the first number , the last of the second and the last of the third. If they are looked a little bit rationally and attentive this solutions are : 6 , (2 , 3) (A) ; 4, (2 , 7) (B) and 2 , (6 , 7) (C) . This simplify things a lot because it is implied that in order for a solution to be that solution must respect the condition that the last numbers can be calculated by multiplication therefore meaning that it is enough in order to verify if there is a solution firstly what are solutions for this smaller case which saves a lot of brute force search sometimes , as here , though not necessarily I think. Now the second term of third number in order of the cases where m is the last term of the multiplication of the first and the decimal unit of the second will be either m + 0 (A) , m + 1(B) or m + 4 (C) but for case A as 2 and 3 will both if placed as the first number multiplied with the decimal unit of the second term produce a number which either is 1 or 2 this case can be eliminated since 1 or 2 will end in the place of the hundred digit of the third which can't be therefore remaining just with case B and C. For case B the numbers that remain to be placed are 3 , 5 and 6 and for case C they are 4 , 3 and 5 and after some rational considerations the solutions come out.

A general way by which such problems can be solved would be interesting nonetheless but that would imply a rational understanding of the way under such calculation things work , an understanding of the principles which will of course not take cases haphazard and trying countless possibilities , and would imply maybe some use of modular arithmetic , anyone have any ideas if there is such a general understanding of it ?