BIDONGYU

Logic Level 4

B I D O N G + Y U 2 0 1 7 J I A Y O U \begin{array}{cccccc}&&B&I&D&O&N&G\\+&&Y&U&2&0&1&7\\ \hline &&J&I&A&Y&O&U \end{array}

Given that each alphabet in the cryptogram above represents distinct digits.

Find the number of possible solutions to the cryptogram.


The answer is 4.

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2 solutions

Donglin Loo
Sep 26, 2018

Look at the second column from left.

We know that the carry over is at most 1 1 .( See the theorem in 3. )

We deduce that U=0 \textbf{U=0} or U=9 \textbf{U=9}

Case 1:U=0 \textbf{Case 1:U=0}

B I D O N G + Y 0 2 0 1 7 J I A Y O 0 \begin{array}{cccccc}&&B&I&D&O&N&G\\+&&Y&0&2&0&1&7\\ \hline &&J&I&A&Y&O&0 \end{array}

We deduce that G=3 \textbf{G=3} .

B I D O N 3 + Y 0 2 0 1 7 J I A Y O 0 \begin{array}{cccccc}&&B&I&D&O&N&3\\+&&Y&0&2&0&1&7\\ \hline &&J&I&A&Y&O&0 \end{array}

We see that there is a carry over at the third column from right.

We deduce that N=8 \textbf{N=8} or N=9 \textbf{N=9}

When N=8 \textbf{N=8} ,

B I D O 8 3 + Y 0 2 0 1 7 J I A Y O 0 \begin{array}{cccccc}&&B&I&D&O&8&3\\+&&Y&0&2&0&1&7\\ \hline &&J&I&A&Y&O&0 \end{array}

O=0 \textbf{O=0} , which contradicts with U=0 \textbf{U=0}

N=9 \therefore \textbf{N=9}

B I D O 9 3 + Y 0 2 0 1 7 J I A Y O 0 \begin{array}{cccccc}&&B&I&D&O&9&3\\+&&Y&0&2&0&1&7\\ \hline &&J&I&A&Y&O&0 \end{array}

O=1 \textbf{O=1}

B I D 1 9 3 + Y 0 2 0 1 7 J I A Y 1 0 \begin{array}{cccccc}&&B&I&D&1&9&3\\+&&Y&0&2&0&1&7\\ \hline &&J&I&A&Y&1&0 \end{array}

Y=2 \textbf{Y=2}

B I D 1 9 3 + 2 0 2 0 1 7 J I A 2 1 0 \begin{array}{cccccc}&&B&I&D&1&9&3\\+&&2&0&2&0&1&7\\ \hline &&J&I&A&2&1&0 \end{array}

The remaining unused digits are 4,5,6,7,8 \textbf{4,5,6,7,8} .

(B,I,D,J,A) = (4,8,5,6,7) , (5,8,4,7,6) , (5,4,6,7,8) , (6,4,5,8,7) \textbf{(B,I,D,J,A)}=\textbf{(4,8,5,6,7)},\textbf{(5,8,4,7,6)},\textbf{(5,4,6,7,8)},\textbf{(6,4,5,8,7)}

There are a total number of 4 \textbf{4} solutions when U=0 \textbf{U=0} .

Case 2: U=9 \textbf{Case 2: U=9}

B I D O N G + Y 9 2 0 1 7 J I A Y O 9 \begin{array}{cccccc}&&B&I&D&O&N&G\\+&&Y&9&2&0&1&7\\ \hline &&J&I&A&Y&O&9 \end{array}

We deduce that G=2 \textbf{G=2} .

B I D O N 2 + Y 9 2 0 1 7 J I A Y O 9 \begin{array}{cccccc}&&B&I&D&O&N&2\\+&&Y&9&2&0&1&7\\ \hline &&J&I&A&Y&O&9 \end{array}

Now, let's look at third column, O \textbf{O} and Y \textbf{Y} are different digits so the carry over is 1 1 .

N+1 \textbf{N+1} 10 \geq10\Rightarrow N \textbf{N} 9 \geq9

U=9 \textbf{U=9} and therefore value of N \textbf{N} does not exist.

U=9 \textbf{U=9} is unsuitable.

We conclude that there are 4 \boxed{4} solutions to the cryptogram.

You can solve this using code:

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from itertools import permutations
count=0
for b,i,d,o,n,g,y,u,j,a in permutations('0123456789',10):
  if int(b+i+d+o+n+g)+int(y+u+'2017')==int(j+i+a+y+o+u):
    count+=1
print(count)

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Solution:
4

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