Solve the cryptarithms given below:

Question

Homi · Accepted Answer

We will solve each cryptarithm by analyzing the column additions and using logical deduction to find the unique digit represented by each letter.

Step 1 — Solving Cryptarithm (i) Let us analyze the addition problem: A1

1B B0

First, consider the units column: $1 + B$ must result in a number ending in 0. Since B is a digit from 0 to 9, $1+B$ can be 10. $1 + B = 10$ $B = 10 - 1$

$\boxed{B = 9}$ This means there is a carry-over of 1 to the tens column.

Next, consider the tens column: The sum of A, 1, and the carry-over (1) must equal B. $A + 1 + 1 = B$ $A + 2 = B$ We already found $B=9$ . Let us substitute this value into the equation: $A + 2 = 9$ $A = 9 - 2$

$\boxed{A = 7}$ Let us verify the solution: 71

19 90 The solution is consistent.

Step 2 — Solving Cryptarithm (ii) Let us analyze the addition problem: AB

37 6A

First, consider the units column: $B + 7$ must result in a number ending in A. This means $B + 7 = A$ or $B + 7 = A + 10$ . If $B+7 = A$ , there is no carry-over to the tens column. If $B+7 = A+10$ , there is a carry-over of 1 to the tens column.

Next, consider the tens column: $A + 3 + \text{carry} = 6$ . If there is no carry-over from the units column (carry = 0): $A + 3 + 0 = 6$ $A = 6 - 3$ $A = 3$ If $A=3$ , then from the units column $B+7=A \Rightarrow B+7=3$ . This is impossible for a non-negative digit B. So, there must be a carry-over of 1 from the units column.

Now, assume there is a carry-over of 1 from the units column: From the tens column: $A + 3 + 1 = 6$ $A + 4 = 6$ $A = 6 - 4$

$\boxed{A = 2}$ Now, use this value of A in the units column equation with a carry-over of 1: $B + 7 = A + 10$ $B + 7 = 2 + 10$ $B + 7 = 12$ $B = 12 - 7$ $\boxed{B = 5}$ Let us verify the solution: 25

37 62 The solution is consistent.

Step 3 — Solving Cryptarithm (iii) Let us analyze the addition problem: ON

ON PO

This means $2 \times ON = PO$ . First, consider the units column: $N + N$ must result in a number ending in O. So, $2N = O$ or $2N = O + 10$ . Let $c_1$ be the carry-over to the tens column. So $c_1$ is 0 if $2N=O$ , and 1 if $2N=O+10$ .

Next, consider the tens column: $O + O + c_1$ must result in P. So, $2O + c_1 = P$ . Since PO is a two-digit number, P cannot be 0. Also, O cannot be 0 as it's the leading digit of ON. The letters O, N, P must represent distinct digits.

Let's consider the case where $c_1 = 0$ : This means $2N = O$ . So $N = O/2$ . Also, $2O = P$ . Since $N$ must be an integer, $O$ must be an even digit. Possible values for O (excluding 0): 2, 4, 6, 8.

If $O=2$ : $N = 2/2 = 1$ . $P = 2 \times 2 = 4$ . The digits are $O=2, N=1, P=4$ . These are distinct. Let us verify: 21

21 42 This is a valid solution.

(Note: There are other valid solutions for this cryptarithm, but we provide one.)

$\boxed{O=2, N=1, P=4}$

Step 4 — Solving Cryptarithm (iv) Let us analyze the addition problem: QR

QR PRR

This means $2 \times QR = PRR$ . First, consider the units column: $R + R$ must result in a number ending in R. So, $2R = R$ or $2R = R + 10$ . If $2R = R$ : $R = 0$ If $2R = R + 10$ : $R = 10$ Since R must be a single digit, $R=10$ is not possible.

$\boxed{R = 0}$ If $R=0$ , then $2R=0$ . So there is no carry-over to the tens column (carry = 0).

Next, consider the tens column: $Q + Q + \text{carry}_1$ must result in a number ending in R. We know $R=0$ and $\text{carry}_1 = 0$ . So, $2Q + 0 = 0$ or $2Q + 0 = 0 + 10$ . If $2Q = 0$ : $Q = 0$ However, Q and R must represent distinct digits. Since $R=0$ , $Q$ cannot be 0. So, $2Q = 10$ . $Q = 10 / 2$

$\boxed{Q = 5}$ This means there is a carry-over of 1 to the hundreds column (carry = 1).

Finally, consider the hundreds column: The carry-over from the tens column must be P. So, $P = \text{carry}_2$ . We found $\text{carry}_2 = 1$ .

$\boxed{P = 1}$ Let us verify the solution: 50

50 100 Here, $Q=5, R=0, P=1$ . The result PRR is 100. All letters $Q, R, P$ represent distinct digits (5, 0, 1). The solution is consistent.

Answer

(i) $A=7, B=9$ (ii) $A=2, B=5$ (iii) $O=2, N=1, P=4$ (iv) $Q=5, R=0, P=1$

Question 39

Answer

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