CA.....B is no more CA.....B!

Logic Level 3

C A A A A A B × D D D D D D B B B B 7 C C C 2 \begin{array} { l l l l } & & & {\color{#3D99F6}C } & {\color{#D61F06}A }&{\color{#D61F06} A} &{\color{#D61F06} A } & \cdots &{\color{#D61F06} A} &{\color{#D61F06} A }&{\color{#20A900} B } \\ & \times & & & {\color{#BA33D6}D }& {\color{#BA33D6}D} & {\color{#BA33D6}D} & \cdots & {\color{#BA33D6}D} & {\color{#BA33D6}D} &{\color{#BA33D6} D }\\ \hline {\color{#20A900}B} &{\color{#20A900} B } & \cdots & {\color{#20A900} B} &{\color{#20A900} B} & 7 &{\color{#3D99F6} C }& \cdots & {\color{#3D99F6}C }&{\color{#3D99F6}C} & 2 \\ \end{array} In the above cryptogram , each letter represents non-zero digits where the number of A , B , C {\color{#D61F06} A}, {\color{#20A900}B },{\color{#3D99F6} C} is 9 9 and D {\color{#BA33D6} D} is 10 10 .

Find the sum of all the digits of the obtained product.

This is original problem (after being inspired)


Clarifications: If 123 × 654 = 80442 123\times654 = 80442 then 8 + 0 + 4 + 4 + 2 = 18 8+0+4+4+2= 18 is sum of all digits of the obtained product.

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1 solution

Hana Wehbi
May 26, 2018

Nice problem. I am still thinking of a nice solution.

Glad that you like the problem. I wish to see your solution too. :) :)

Naren Bhandari - 3 years ago

1 pending report

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