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Logic Level 2

A B C D × 9 D C B A \begin{array}{ccccc} & &A & B & C&D\\ \times & & & & &9 \\ \hline & & D & C & B &A\\ \hline \end{array}

If A , B , C A,B,C and D D are distinct digits that satisfy the cryptogram above, find A + B + C + D A+B+C+D .

Note : The leading digits cannot be zero.


This problem is a part of my set Numbers in Disguise - Cryptarithms .


The answer is 18.

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5 solutions

Tran Quoc Dat
Feb 22, 2015

First, we all know A = 1 , D = 9 \mathsf{A=1, D=9} .

1 B C 9 = 1009 + B C × 10 \overline{\mathsf{1BC9}} = 1009 + \overline{\mathsf{BC}} \times 10 .

1 B C 9 × 9 = 9081 + B × 900 + C × 90 \overline{\mathsf{1BC9}} \times 9 = 9081 + \mathsf{B} \times 900 + \mathsf{C} \times 90 .

9 C B 1 = 9001 + B × 10 + C × 100 \overline{\mathsf{9CB1}} = 9001 + \mathsf{B} \times 10 + \mathsf{C} \times 100 .

Now we have this equation:

9081 + B × 900 + C × 90 = 9001 + B × 10 + C × 100 9081 + \mathsf{B} \times 900 + \mathsf{C} \times 90 = 9001 + \mathsf{B} \times 10 + \mathsf{C} \times 100 .

B × 890 + 80 = C × 10 \mathsf{B} \times 890 + 80 = \mathsf{C} \times 10 .

B × 89 + 8 = C \mathsf{B} \times 89 + 8 = \mathsf{C} .

If C 8 \mathsf{C} \not = 8 , there's no solution for B \mathsf{B} .

If C = 8 \mathsf{C} = 8 then B = 0 \mathsf{B} = 0 .

A + B + C + D = 1 + 0 + 8 + 9 = 18 \mathsf{A + B + C + D} = 1 + 0 + 8 + 9 = \boxed{18} .

Hung Woei Neoh
Apr 14, 2016

Here's a step by step solution:

[ b ] r A B C D × 9 [ b ] r D C B A \frac{\begin{array}{c}[b]{r} ABCD\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\;\; DCBA\\ \hline \end{array}}

Now, we know that A 0 A \neq 0 . If A > 1 A>1 , the product will be a 5 digit number, and that does not satisfy the cryptogram. From here, we can confirm that A = 1 A = 1

From here, we know that D × 9 D \times 9 gives us a number that ends with 1 1 . Looking at the multiplication table, only 9 × 9 = 81 9 \times 9 = 81 satisfies the requirement. Therefore,

D = 9 D = 9

Substitute these two numbers into the cryptogram, and we get:

[ b ] r 1 B C 9 × 9 [ b ] r 9 C B 1 \frac{\begin{array}{c}[b]{r} 1BC9\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\quad\; 9CB1\\ \hline \end{array}}

Now, we convert it into an equation as below:

( 1009 + 100 B + 10 C ) ( 9 ) = 9001 + 100 C + 10 B (1009 + 100B + 10C)(9) = 9001 + 100C + 10B

Simplifying this equation gives:

89 B + 8 = C 89B + 8 = C

Now, remember that the values of B B and C C are integers for a digit, which gives:

0 B 9 0 C 9 0 \leq B \leq 9 \quad\quad 0 \leq C \leq 9

If B > 0 B>0 , C C will definitely be larger than 9 9 , and that does not satisfy the cryptogram.

Therefore, the only solution is B = 0 B = 0 , and C = 89 ( 0 ) + 8 = 8 C = 89(0) + 8 = 8 .

The solved cryptogram is

[ b ] r 1089 × 9 [ b ] r 9801 \frac{\begin{array}{c}[b]{r} 1089\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\quad\;\; 9801\\ \hline \end{array}}

Calculate the final answer: A + B + C + D = 1 + 0 + 8 + 9 = 18 A + B + C + D = 1 + 0 + 8 + 9 = \boxed{18}

Well explained :).

Mthokozisi Kabvala - 4 years, 10 months ago
Alex Li
Mar 26, 2016

If ABCD is greater than 1111, then the solution will be a 5 digit number, and thus be null.

Therefore, A must equal 1, and B 0. Therefore, 1000<ABCD<1111. Any of these values multiplied by 9 give 9 as the leading coefficient.

From here, C is easy to solve for, by doing multiplication we can see that the 8 will carry over from D*9, giving 8+9C = 10C+0, C = 8.

(Sorry, that last bit was poorly worded)

A A
Mar 15, 2016

The four digit number when multiplied by 9 gives a 4 digit number which means that A must be 1 and since the result of A * 9=D , D must be 9. Since the thousand digit of the result is the result of A * 9 plus the result of the decimal digit of B * 9 B must be 0 and since D * 9 = 9 * 9=81 it means that the decimal digit of the result that is B which is 0 is the result of 8 to which we add the unit digit of the product C*9 this being 2 therefore being that C is 8 which means that A = 1 , B = 0 , C = 8 , D = 9 which added gives 18.

Ameya Salankar
May 1, 2014

The numbers are: A = 1 , B = 0 , C = 8 , D = 9 \mathsf{A}=1, \mathsf{B}=0, \mathsf{C}=8, \mathsf{D}=9 .

The cryptarithm was 1089 × 9 = 9801 1089 \times 9 = 9801 .

/ Here is the c program which gives you the answer.TRY it on online compiler. http://www.compileonline.com/compile_c_online.php Code is..................................................................................................................... /

include<stdio.h>

int reverse(int a) {int rev=0,m; m=a; while(m>0) { rev=10*rev+m%10; m=m/10; } return rev; }

void main() {int i; for(i=1;i<=9999;i++) if(reverse(i)==9*i) printf("The answer is %d.",i);

}

Aamir Faisal Ansari - 7 years, 1 month ago

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