Skip to main content
Puzzling

Return to Answer

added 897 characters in body
Source Link
Bubbler
Bubbler
  • 17.1k
  • 1
  • 31
  • 109

 

Plugging in $b=1$ to $L=x+b-1$, we simply get $L=x$, and we can write $IJ \times K = LIQ$. Under the constraint that no digits in the equation can be 0 or 9 and the five digits $JKLIQ$ are unique, I decided to brute-force the equation and found four candidates: $$34 \times 7 = 238, 67 \times 4 = 268, 72 \times 8 = 576, 83 \times 7 = 581$$ With two additional conditions that $O=K-1$ and $P=10-K+L$ are also unique, only the first one survives, and gives the final answer.

EDIT: if you start with $O=K-1$ and $P=10-K+L$ combined with $\{O,P,J,K,L,I,Q\} \subset \{1,2,3,4,5,6,7,8\}$, you can actually work out the brute force by hand:

K = 2: O = 1, P = 8+L, impossible
K = 3: O = 2, P = 7+L, PL = 81, IJ×3 = 1IQ, {IJQ}⊂{4567} but no Q for any J
K = 4: O = 3, P = 6+L, PL = 71 or 82
IJ×4 = 1IQ, {IJQ}⊂{2568}: I=2, impossible
IJ×4 = 2IQ, {IJQ}⊂{1567}: Q=6, impossible
K = 5: O = 4, P = 5+L, PL = 61, 72, or 83
IJ×5 = LIQ: impossible
K = 6: O = 5, P = 4+L, PL = 73 or 84
IJ×6 = 3IQ, {IJQ}⊂{1248}: impossible
IJ×6 = 4IQ, {IJQ}⊂{1237}: Q=2, J=7, impossible
K = 8: O = 7, P = 2+L, PL = 31, 42, 53, 64
IJ×8 = LIQ: 4J×8 = 34Q, {JQ}⊂{126}, impossible
K = 7: O = 6, P = 3+L, PL = 41, 52, or 85
IJ×7 = 1IQ: impossible
IJ×7 = 2IQ: 3J×7 = 23Q, 34×7 = 238
IJ×7 = 5IQ: I>=8 but P=8, so impossible

Plugging in $b=1$ to $L=x+b-1$, we simply get $L=x$, and we can write $IJ \times K = LIQ$. Under the constraint that no digits in the equation can be 0 or 9 and the five digits $JKLIQ$ are unique, I decided to brute-force the equation and found four candidates: $$34 \times 7 = 238, 67 \times 4 = 268, 72 \times 8 = 576, 83 \times 7 = 581$$ With two additional conditions that $O=K-1$ and $P=10-K+L$ are also unique, only the first one survives, and gives the final answer.

 

Plugging in $b=1$ to $L=x+b-1$, we simply get $L=x$, and we can write $IJ \times K = LIQ$. Under the constraint that no digits in the equation can be 0 or 9 and the five digits $JKLIQ$ are unique, I decided to brute-force the equation and found four candidates: $$34 \times 7 = 238, 67 \times 4 = 268, 72 \times 8 = 576, 83 \times 7 = 581$$ With two additional conditions that $O=K-1$ and $P=10-K+L$ are also unique, only the first one survives, and gives the final answer.

EDIT: if you start with $O=K-1$ and $P=10-K+L$ combined with $\{O,P,J,K,L,I,Q\} \subset \{1,2,3,4,5,6,7,8\}$, you can actually work out the brute force by hand:

K = 2: O = 1, P = 8+L, impossible
K = 3: O = 2, P = 7+L, PL = 81, IJ×3 = 1IQ, {IJQ}⊂{4567} but no Q for any J
K = 4: O = 3, P = 6+L, PL = 71 or 82
IJ×4 = 1IQ, {IJQ}⊂{2568}: I=2, impossible
IJ×4 = 2IQ, {IJQ}⊂{1567}: Q=6, impossible
K = 5: O = 4, P = 5+L, PL = 61, 72, or 83
IJ×5 = LIQ: impossible
K = 6: O = 5, P = 4+L, PL = 73 or 84
IJ×6 = 3IQ, {IJQ}⊂{1248}: impossible
IJ×6 = 4IQ, {IJQ}⊂{1237}: Q=2, J=7, impossible
K = 8: O = 7, P = 2+L, PL = 31, 42, 53, 64
IJ×8 = LIQ: 4J×8 = 34Q, {JQ}⊂{126}, impossible
K = 7: O = 6, P = 3+L, PL = 41, 52, or 85
IJ×7 = 1IQ: impossible
IJ×7 = 2IQ: 3J×7 = 23Q, 34×7 = 238
IJ×7 = 5IQ: I>=8 but P=8, so impossible

Source Link
Bubbler
Bubbler
  • 17.1k
  • 1
  • 31
  • 109

The answer is

         7
    ______
934 ) 7210
      6538
      ----
       672

(H, I, J, K, L, M, N, O, P, Q) = (9, 3, 4, 7, 2, 1, 0, 6, 5, 8)

Some reasoning (for the equations, $KL$ denotes digit concatenation; all multiplications use explicit $\times$):

$L \ne 0$, since otherwise $N=Q$ from the lowest digit. Let's say $KL + IQ = bMN$ where $b\in\{0,1\}$. Then $OP + O + b = KL$. We know that $K = O + 1$ and $L \ge 1$, so $$OP + O + b = 10\times(K-1) + P + O + b = KL = 10\times K + L \\ P + O + b = L + 10 \\ P + O = L + 10 - b \ge 10$$ If we plug in various values of $K$ into $HIJ \times K > OP00$, we get $$HIJ \times 2 > 1900, HIJ \times 3 > 2800, \dots, HIJ \times 8 > 7300$$ and in all cases, $H=9$ and $I\ge 1$.

Now let's say $IJ \times K = xIQ$ where $1 \le x \le I$. ($x$ cannot be 0 because $I$ is nonzero and $K > 1$.) Then we can say $OP = H \times K + x = 9 \times K + x$. If we plug it back into $OP + O + b = KL$, we get $$9\times K + x + (K - 1) + b = KL \\ L = x + b - 1$$

At this point, let's assume $b=0$. As Ross Millikan pointed out (and we found $I\ne 0$), the zero must be either $M$ or $N$. But if $b=0$, $KL + IQ = MN$, so $M\ne 0$ and subsequently $N=0$. We can find out the following: $$L = x-1 \\ KL + IQ = M0 \Rightarrow L+Q=10, K+I+1=M \\ L = x-1 \Rightarrow x+Q=11 \\ Q \le 8 \Rightarrow x \ge 3 \\ M \le 8 \Rightarrow K+I \le 7 $$ Now, $IJ \times K = xIQ > 300$, but any choice of $K$ and $I$ subject to $K+I \le 7$ cannot satisfy $IJ \times K > 300$. Therefore $b=1$.

Plugging in $b=1$ to $L=x+b-1$, we simply get $L=x$, and we can write $IJ \times K = LIQ$. Under the constraint that no digits in the equation can be 0 or 9 and the five digits $JKLIQ$ are unique, I decided to brute-force the equation and found four candidates: $$34 \times 7 = 238, 67 \times 4 = 268, 72 \times 8 = 576, 83 \times 7 = 581$$ With two additional conditions that $O=K-1$ and $P=10-K+L$ are also unique, only the first one survives, and gives the final answer.