Question 41

We will solve each cryptarithmetic puzzle by converting it into an algebraic equation and using logical deduction based on digit properties.

Step 1 — Solving UT × 3 = PUT

Let us represent the numbers using place values. The number UT means $10U + T$. The number PUT means $100P + 10U + T$. So, the equation becomes:

$(10U + T) \times 3 = 100P + 10U + T$

$30U + 3T = 100P + 10U + T$

$20U + 2T = 100P$

$10U + T = 50P$

We know that U, T, P are unique digits. Also, U and P cannot be 0 because they are leading digits. From $10U + T = 50P$: The left side, $10U + T$, is the number UT. Since UT is a two-digit number, $10 \le UT \le 99$. So, $10 \le 50P \le 99$. Dividing by 50, we get $10/50 \le P \le 99/50$. $0.2 \le P \le 1.98$ Since P must be an integer digit, P = 1.

Now substitute P = 1 into the equation: $10U + T = 50 \times 1$

$10U + T = 50$ This means the number UT is 50. So, U = 5 and T = 0. Let us check if U, T, P are unique and valid: U = 5, T = 0, P = 1. All are unique digits. U is not 0, P is not 0. Let's verify the original equation: $50 \times 3 = 150$. Here, UT = 50, PUT = 150. This matches.

$\boxed{\text{U=5, T=0, P=1}}$

Step 2 — Solving AB × 5 = BC

Let us represent the numbers using place values. The number AB means $10A + B$. The number BC means $10B + C$. So, the equation becomes:

$(10A + B) \times 5 = 10B + C$

$50A + 5B = 10B + C$

$50A = 5B + C$ We know that A, B, C are unique digits. Also, A and B cannot be 0 because they are leading digits. The product BC is a two-digit number. This means $AB \times 5 < 100$. So, $AB < 100/5$. $AB < 20$ Since A is the tens digit of AB, A can only be 1. So, A = 1.

Now substitute A = 1 into the equation: $50 \times 1 = 5B + C$

$50 = 5B + C$ The units digit of any number multiplied by 5 must be either 0 or 5. So, C must be 0 or 5. If C = 0: $50 = 5B + 0$

$5B = 50$

$B = 10$ This is not possible, as B must be a single digit. So, C cannot be 0.

If C = 5: $50 = 5B + 5$

$5B = 50 - 5$

$5B = 45$

$B = 9$ So, B = 9 and C = 5. Let us check if A, B, C are unique and valid: A = 1, B = 9, C = 5. All are unique digits. A is not 0, B is not 0. Let's verify the original equation: $19 \times 5 = 95$. Here, AB = 19, BC = 95. This matches.

$\boxed{\text{A=1, B=9, C=5}}$

Step 3 — Solving L2N × 2 = 2NP

Let us represent the numbers using place values. The number L2N means $100L + 20 + N$. The number 2NP means $200 + 10N + P$. So, the equation becomes:

$(100L + 20 + N) \times 2 = 200 + 10N + P$

$200L + 40 + 2N = 200 + 10N + P$ We know that L, N, P are unique digits. L cannot be 0. Also, L, N, P cannot be 2, as 2 is a given digit in the numbers. Let's analyze the multiplication column by column: L 2 N x 2

2 N P From the hundreds column: $L \times 2$ (plus any carry from the tens column) must result in 2 (the leading digit of 2NP). Since L is a single digit and L cannot be 0, the only way $2L + \text{carry}$ can start with 2 is if $L=1$. If $L=1$, then $2 \times 1 + \text{carry} = 2$. This means the carry from the tens column must be 0. So, L = 1.

Now consider the tens column: $2 \times 2 + \text{carry from units column} = N$. Since the carry to the hundreds column is 0, $2 \times 2 + \text{carry from units column}$ must be less than 10. Let $c_1$ be the carry from the units column ($N \times 2$). $4 + c_1 = N$ The carry $c_1$ can be 0 or 1 (since $N \times 2$ can be at most $9 \times 2 = 18$). If $c_1 = 0$: $N = 4 + 0$

$N = 4$ If $N=4$, then $N \times 2 = 4 \times 2 = 8$. The carry $c_1$ is indeed 0. This is consistent. Now we find P from the units column: $N \times 2$ ends in P. $4 \times 2 = 8$ So, P = 8. This gives us L=1, N=4, P=8. These are unique digits (1, 4, 8) and none of them is 2. This is a valid solution.

If $c_1 = 1$: $N = 4 + 1$

$N = 5$ If $N=5$, then $N \times 2 = 5 \times 2 = 10$. The carry $c_1$ is indeed 1. This is consistent. Now we find P from the units column: $N \times 2$ ends in P. $5 \times 2 = 10$ So, P = 0. This gives us L=1, N=5, P=0. These are unique digits (1, 5, 0) and none of them is 2. This is also a valid solution.

There are two possible solutions for this puzzle.

$\boxed{\text{L=1, N=4, P=8 \quad OR \quad L=1, N=5, P=0}}$

Step 4 — Solving XY × 4 = ZX

Let us represent the numbers using place values. The number XY means $10X + Y$. The number ZX means $10Z + X$. So, the equation becomes:

$(10X + Y) \times 4 = 10Z + X$

$40X + 4Y = 10Z + X$

$39X + 4Y = 10Z$ We know that X, Y, Z are unique digits. Also, X and Z cannot be 0 because they are leading digits. The product ZX is a two-digit number. This means $XY \times 4 < 100$. So, $XY < 100/4$. $XY < 25$ Since X is the tens digit of XY, X can only be 1 or 2.

Let's look at the units digit of the multiplication: $Y \times 4$ must end in X. Possible pairs $(Y, X)$ where $4Y$ ends in $X$:

• If $Y=1$, $4Y=4$, so $X=4$. (Not possible, as X must be 1 or 2).
• If $Y=2$, $4Y=8$, so $X=8$. (Not possible, as X must be 1 or 2).
• If $Y=3$, $4Y=12$, so $X=2$. (Possible, $X=2$).
• If $Y=4$, $4Y=16$, so $X=6$. (Not possible, as X must be 1 or 2).
• If $Y=5$, $4Y=20$, so $X=0$. (Not possible, as X cannot be 0).
• If $Y=6$, $4Y=24$, so $X=4$. (Not possible, as X must be 1 or 2).
• If $Y=7$, $4Y=28$, so $X=8$. (Not possible, as X must be 1 or 2).
• If $Y=8$, $4Y=32$, so $X=2$. (Possible, $X=2$).
• If $Y=9$, $4Y=36$, so $X=6$. (Not possible, as X must be 1 or 2).

So, X must be 2. Now we have two cases for Y: Y=3 or Y=8.

Case 1: X = 2, Y = 3. Let's check the original multiplication: $XY = 23$. $23 \times 4 = 92$ So, $ZX = 92$. This means Z = 9 and X = 2. The digits are X=2, Y=3, Z=9. These are unique, and X, Z are not 0. This is a valid solution.

Case 2: X = 2, Y = 8. Let's check the original multiplication: $XY = 28$. $28 \times 4 = 112$ This product is a three-digit number, but ZX must be a two-digit number. So, this case is not possible.

Thus, the only solution is X=2, Y=3, Z=9.

$\boxed{\text{X=2, Y=3, Z=9}}$

Step 5 — Solving PP × QQ = PRP

Let us represent the numbers using place values. The number PP means $10P + P = 11P$. The number QQ means $10Q + Q = 11Q$. The number PRP means $100P + 10R + P = 101P + 10R$. So, the equation becomes:

$(11P) \times (11Q) = 101P + 10R$

$121PQ = 101P + 10R$ We know that P, Q, R are unique digits. Also, P and Q cannot be 0 because they are leading digits. The product PRP is a three-digit number. This means $100 \le PRP \le 999$. So, $100 \le 121PQ \le 999$. Dividing by 121: $100/121 \le PQ \le 999/121$

$0.82 \le PQ \le 8.25$ So, the product of digits P and Q must be an integer between 1 and 8 (inclusive). Also, the first digit of $PRP$ is $P$. This means the first digit of $121PQ$ must be $P$.

Let's test possible values for $PQ$:

• If $PQ = 1$: $121 \times 1 = 121$. So $PRP = 121$. This means $P=1, R=2$.
  • For $PQ=1$, the only possible pair of unique non-zero digits is $(P,Q) = (1,1)$, but P and Q must be unique. So this is not possible.
• If $PQ = 2$: $121 \times 2 = 242$. So $PRP = 242$. This means $P=2, R=4$.
  • For $PQ=2$, possible pairs are $(P,Q) = (1,2)$ or $(2,1)$.
  • If $(P,Q) = (1,2)$, then $P=1$. But from $PRP=242$, $P=2$. This is a contradiction.
  • If $(P,Q) = (2,1)$, then $P=2$. This matches $P=2$ from $PRP=242$.
    • Digits are P=2, Q=1, R=4. These are unique, and P, Q are not 0. This is a valid solution.
• If $PQ = 3$: $121 \times 3 = 363$. So $PRP = 363$. This means $P=3, R=6$.
  • For $PQ=3$, possible pairs are $(P,Q) = (1,3)$ or $(3,1)$.
  • If $(P,Q) = (1,3)$, then $P=1$. But from $PRP=363$, $P=3$. This is a contradiction.
  • If $(P,Q) = (3,1)$, then $P=3$. This matches $P=3$ from $PRP=363$.
    • Digits are P=3, Q=1, R=6. These are unique, and P, Q are not 0. This is a valid solution.
• If $PQ = 4$: $121 \times 4 = 484$. So $PRP = 484$. This means $P=4, R=8$.
  • For $PQ=4$, possible pairs are $(P,Q) = (1,4)$, $(4,1)$, or $(2,2)$. $(2,2)$ is not allowed as P and Q must be unique.
  • If $(P,Q) = (1,4)$, then $P=1$. But from $PRP=484$, $P=4$. This is a contradiction.
  • If $(P,Q) = (4,1)$, then $P=4$. This matches $P=4$ from $PRP=484$.
    • Digits are P=4, Q=1, R=8. These are unique, and P, Q are not 0. This is a valid solution.
• If $PQ = 5$: $121 \times 5 = 605$. So $PRP = 605$. This means $P=6, R=0$.
  • For $PQ=5$, possible pairs are $(P,Q) = (1,5)$ or $(5,1)$.
  • If $(P,Q) = (1,5)$, then $P=1$. But from $PRP=605$, $P=6$. Contradiction.
  • If $(P,Q) = (5,1)$, then $P=5$. But from $PRP=605$, $P=6$. Contradiction.
• If $PQ = 6$: $121 \times 6 = 726$. So $PRP = 726$. This means $P=7, R=2$.
  • For $PQ=6$, possible pairs are $(1,6), (6,1), (2,3), (3,2)$. None of these pairs has $P=7$. Contradiction.
• If $PQ = 7$: $121 \times 7 = 847$. So $PRP = 847$. This means $P=8, R=4$.
  • For $PQ=7$, possible pairs are $(1,7), (7,1)$. None of these pairs has $P=8$. Contradiction.
• If $PQ = 8$: $121 \times 8 = 968$. So $PRP = 968$. This means $P=9, R=6$.
  • For $PQ=8$, possible pairs are $(1,8), (8,1), (2,4), (4,2)$. None of these pairs has $P=9$. Contradiction.

There are three possible solutions for this puzzle.

$\boxed{\text{P=2, Q=1, R=4 \quad OR \quad P=3, Q=1, R=6 \quad OR \quad P=4, Q=1, R=8}}$

Step 6 — Solving JK × 6 = KKK

Let us represent the numbers using place values. The number JK means $10J + K$. The number KKK means $100K + 10K + K = 111K$. So, the equation becomes:

$(10J + K) \times 6 = 111K$ We know that J, K are unique digits. Also, J and K cannot be 0 because they are leading digits. Let's simplify the equation: $60J + 6K = 111K$

$60J = 111K - 6K$

$60J = 105K$ We can divide both sides by their greatest common divisor, which is 15. $60J / 15 = 105K / 15$

$4J = 7K$ Since 4 and 7 are coprime (they share no common factors other than 1), J must be a multiple of 7, and K must be a multiple of 4. J and K are single digits from 0 to 9. Also, J and K cannot be 0. If J is a multiple of 7:

• If J = 7: $4 \times 7 = 7K$

  $28 = 7K$

  $K = 4$ So, J=7 and K=4. These are unique digits, and neither is 0. This is a valid solution.

• If J is any other multiple of 7 (e.g., 14), it would be a two-digit number, which is not allowed.

If K is a multiple of 4:

• If K = 4: (already found above)

• If K = 8: $4J = 7 \times 8$

  $4J = 56$

  $J = 14$ This is a two-digit number, which is not allowed.

• If K is any other multiple of 4 (e.g., 12), it would be a two-digit number.

Thus, the only solution is J=7, K=4. Let's verify the original equation: $74 \times 6 = 444$. Here, JK = 74, KKK = 444. This matches.

$\boxed{\text{J=7, K=4}}$

Answer

(i) U=5, T=0, P=1 (ii) A=1, B=9, C=5 (iii) L=1, N=4, P=8 OR L=1, N=5, P=0 (iv) X=2, Y=3, Z=9 (v) P=2, Q=1, R=4 OR P=3, Q=1, R=6 OR P=4, Q=1, R=8 (vi) J=7, K=4