Triangular cryptogram

The greatest distinct 2 digit number subtracted from the smallest distinct 3 digit number is 102 - 98 = 4, which is the smallest value of F. Therefore, F equal to 1, 2, and 3 are not possible. Hence the solution is 4.

Since the subtraction results in a one digit number it is evident that it is $\overline{10C} - \overline{9E} = F$ . As these above digits have already been used, to get the smallest value of F, we let $C = 2$ and $E = 8$ . This means that the smallest possible value of F is 4.

Jus try to make the smallest number with unique A,B,C,D,E. This is clearly $102-98=4$ . Since no smaller number can be made, this eliminates 1,2,3. Therefore the answer should be $\boxed{4}$

102-98=4 is one.

If F is the solution, then A must be 1 and D must be 9. Also, B must be 0.

Now we have 10C - 9E = F

100 - 90 + (C - E) = F

10 + C - E = F

Since digits 1,0, and 9 have been used, and C<E if F is positive, then we can make the following inequalities:

1<C<7

2<E<8

We can then try values, which yield values of F from 4 to 9. 4 is the only valid choice in the problem, so 4 is a possible value.

It holds that A=1, because if A is greater, then it wouldn't be possible to get a single digit result by subtracting two-digit number from A. So, then we get: 100+10B+C=F+10D+E. The right hand side has an upper bound of 105 (D=9, F=9, E=8 for example), so B must be 0 (if B is not zero, then the left side is at least 120, since digits need to be different). We have a lower bound of 102 for the left side. It remains to look at what is possible and what not.

The largest number (ABC) has to be a small 100’s number and the smallest number has to be a large 90’s number. We can’t have 104 and 95, say, because the difference is greater than any of the options. Also, each digit is distinct, as it says in the question, so you couldn’t have 102 and 99 (difference of 3), so 3 can’t be right.

A difference of 2 is not possible as the two numbers would violate the distinct digit rule (102 and 100 say).

A difference of 1 is not possible either for the same reasons as a difference of 2.

This leaves the only possible answer to be 4 (102 and 98).

Each digit is a different number eliminating numbers like 100-99. Smallest 3 digit number you can create with that is 102. You must then subtract a 2 digit number to get a 1 digit number or work the other way around eliminating 1 and 2 because you will get 101 or 100 which repeat numbers. 3 is out when you realize 102-3=99, repeating numbers. Only option left is 4.

We ned ABC to be the smallest 3 digit number, since A .ne. 0. 103 cannot work since subtracting the largest 2 digit number with different digits(98), gives an answer of 5. 101 contains 2 of the same digits, so we settle on ABC = 102, leaving DE to be 98, with a difference of F = 4. Ed Gray

Since $\overline{ABC}-\overline{DE} = F$ $\implies \overline{DE}+F = \overline{ABC}$ . It is obvious that $A=1$ , since the largest $\overline{DE}+F = 98+7 = 105$ and that $D=9$ and $E+F \ge 10$ . Since $A=1$ , $F \ne 1$ . If $F=2$ , $\overline{DE}+F = 98+2 = 100$ , then $B=C$ not acceptable. $F=3$ $\implies 98+3 = 101$ , $A=C$ not acceptable. $F=4$ , $\implies 98+4 = 102$ , all the digits are distinct. Therefore, $F=\boxed{4}$ is acceptable.

One solution:

Note that the minuend is a three digit integer, the subtrahend is a two digit integer and the difference is a one digit integer. So we start with the minuend, it must be the lowest three digit number with distinct digits and that is 102. Then the subtrahend must be the highest two digit number with distinct digits and that is 98. Subtracting 98 from 102 gives a difference of 4.

First we see all answer in above are small so $\overline{ABC}$ have to be 3 digit smallest distinct number & $\overline{DE}$ have to be 2 digit distinct number.Next the smallest 3 digit distinct number of $\overline{ABC}=102$ .After that the biggest 2 digit distinct number of $\overline{DE}=98$ .So,the $F$ is $\color{#3D99F6}\boxed{\large{4}}$

102 - 98

is the extreme case...

Since the 100 digit of the result is 0, A can only be 1.

Therefore B needs to be smaller then D. The difference of B and D as well as a potential carryover need to be 10, so that there is a carryover to cancel the 1 of A and have a 0 for the 10th digit of the result. Since the carryover can only be 0 or 1 the formula is B - D = -10 or B - D - 1 = -10 Since we are working with digits the only possible combination is: B = 0 and D = 9 with a carryover of 1.

Sofar it's 10C - 9E = F

Since we need a carryover for the 10th digit C needs to be bigger then E.

If C = 2 and ... E = 3, then F = 9 which is no destict digit E = 4 or 5, then F = 8 or 7 E = 6, then F = 6 which is no destict digit E = 7 or 8, then F = 5 or 4

There we already have one of the values in the selection. Due to the wording of the question this is enough and we can use 4 as the answer.

I want to complete the possible combinations for C, E and F.

If C and E are only one apart then F is 9 and therefore not destict.

If C is even (2,4,6) there is one combination of E and F where they are not distinct. For all the odd numbers for C as well as the other combinations for even values of C there are distinct combinations for E and F, with the values of 8 down to C + 2.

Since all those combinations will produce values for F between 8 and 4 there won't be any other values from the options.