82 of 100: The Last Cryptograms

Using the multiplication:

AAA = A(111)

A0A0A=A(10101)

BC must be 10101/111=91

Using the addition:

If A+A>10, then A+1=8, so A=7

If A+A<10, then A+1=9, so A=8, but A+A is then >10, contradiction.

Thus, A=7

I was able to solve the cryptogram fairly quickly using the process of elimination.

I. Letter A CAN NOT be 5 since that would make letter D equal to zero in the addition A A + C A . As specified in the problem, all the digits are nonzero digits.

II. Letter A CAN NOT be 8 or 9 because in either case, the minimum value of A A + C A would be a 3-digit number(88+18=106 or 99+19=118).

III. That leaves 6 or 7 as the only possible choices for letter A. We know from the multiplication the last digit of A * C is A. Since all the letters represent different digits, the only choice for letter C which meets all the criteria is C = 1 . We can then decide the digits for B and D when A = 6 or 7 and C = 1 in the addition A A + C A . The two sets of possible values are as follows:

(a) A = 6, C = 1, B = 8, D = 2

(b) A = 7, C = 1, B = 9, D = 4

Plug the digits in (a) into the multiplication and we get 666 * 81 = 53,946, which DOES NOT fit the cryptogram. Plug the digits in (b) into the multiplication and we get 777 * 91 = 70,707, which fits the cryptogram perfectly!

Therefore, the only solution is {A, B, C, D} = {7, 9, 1, 4} and A = $\boxed{7}$ .

$\begin{aligned} \overline{AAA} \times \overline{BC} & = \overline{A0A0A} \\ 111A \times \overline{BC} & = 10101A \\ \implies \overline{BC} & = \frac {10101}{111} = 91 \\ \implies B & = 9 \\ C & = 1 \end{aligned}$

Then, we have:

$\begin{aligned} \overline{AA} + \overline{CA} & = \overline{BD} \\ 12A + 10{\color{#3D99F6}C} & = 10{\color{#3D99F6}B} + D & \small \color{#3D99F6} \text{Note that }B=9, \ C=1 \\ 12A + 10 & = 90 + D \\ {\color{#3D99F6}12}A & = \color{#3D99F6}80 + D & \small \color{#3D99F6} \text{Note that }\overline{8D} \text{ must be a multiple of 12} \\ \implies \overline{8D} & = 84 \\ \implies A & = \frac {84}{12} = \boxed{7} \end{aligned}$

AAA = 111 $\times$ A

A0A0A = 10101 $\times$ A

We get the following equation from the multiplication:

111 $\times$ A $\times$ (10 $\times$ B+C) = 10101 $\times$ A

-> 10 $\times$ B+C=91

Since B represents a nonzero digit, 10 $\times$ B = B0 so B has to be 9 and therefore C hast to be 1

According to the possible solutions A has to be equal to or greater than 5 and A+A is equal or greater than 10

So we can infer from the addition

A+C+1=B

-> A=7

First, let's use the multiplication to figure out the values of B and C. BC = A0A0A/AAA = [A(10101)]/[A(111)] = 10101/111 = 91. This means that B = 9 and C = 1. We now have two possible routes to take. Either A<5 and A+1 = 9, or A is greater than or equal to 5, and A+1+1 = 9. If we look at the first possibility where A+1=9, A=8. However, A>5, which contradicts the assumption the first possibility was made off of. That leaves the second possibility, A+1+1=9, resulting in A=7 .

AAA=111a

A0A0A=10101a

BC=10101a/111a=91

Therefore B=9 and C=1

If A=5,

55+15=70 while B=9, not possible

if A=6,

66+16=82, while B=9, not possible

if A=7,

77+17=94 while B=9 possible answer is & for A

Therefore A=7

From the 1st equation many people have got 111 x (10 X B + C) = 10101, but then by viewing the units column C = 1. Then 111 x (10B + 1) = 10101, gives that 1110 x B = 9990. So B = 9. Then from the 2nd equation if A<5 then there is no 'carry forward 1', and so A + C = B from which we get that A = 8, which is clearly absurd because A had to be less than 5. So A>=5 then 1 + A + C (1) = B (9). So A=7, (if A were 5 then D would be equal to zero which we were told was a non-zero number). From this D=4. And to check 777 x 91 = 70707 indeed! And 77 + 17 = 94 indeed! Regards, David

A check shows that A = 5, 6, 8, or 9 does not work, then A = 7, Since AC has to give you A, then C = 1. The addition shows that D = 4 and B = 9. So. 7 7 7 x 91 = 70707 and,
7 7 + 17 = 9 4 So, A = 7, C = 1, B = 9, and D = 4.

From the multiplication-

(111)A x (10B+C) = (10101)A

10B+C = 91

B=9, C=1 ......................[ As 0 <B,C< 10 ]

From the addition:

11A + 10C + A = 10B + D

12A + 10= 90+D

12A = 80 + D

12A= 84 ........................[ As (80+D) ≡ 0 (mod 12) ; 80+D= 84]

A = 7

From the given addition, A ≠ 5, since D cannot be 0. Also, the minimum and maximum values at this juncture of (A+A) are 12 and 18 respectively, (since A ∈ [6,9]), so the value that gets carried over is always 1. Thus we have:

1 + A + C = B,

Coming to the multiplication, we can write:

BC = $\frac{A0A0A0}{AAA}$ = $\frac{A(101010)}{A(111)}$ =91

Therefore, B = 9 and C = 1, and hence, the value of A can easily be calculated as 7.