Number Play | IT

Question 41

Context: These types of questions can be interesting and fun to solve! Here are some more questions like this for you to try out. Find out what each letter stands for. Share how you thought about each question with your classmates; you may find some new approaches.

Q. Find out what each letter stands for:

(i) $\begin{array}{r} \text{YY} \\ +\quad \text{Z} \\ \hline \text{ZOO} \end{array}$

(ii) $\begin{array}{r} \text{B5} \\ +\quad \text{3D} \\ \hline \text{ED5} \end{array}$

(iii) $\begin{array}{r} \text{KP} \\ +\quad \text{KP} \\ \hline \text{PRR} \end{array}$

(iv) $\begin{array}{r} \text{C1} \\ +\quad \text{C} \\ \hline \text{1FF} \end{array}$

Solution

We need to find the digit each letter stands for. Each letter is a unique digit from 0 to 9.

Step 1 — Solving (i) YY + Z = ZOO

Let us look at the hundreds column. The sum ZOO has three digits. The sum of YY and Z gives ZOO. The carry to the hundreds column must be Z. The maximum sum of $99 + 9$ is $108$ . So, the carry to the hundreds column can only be 1. This means Z must be 1.

$\text{Z} = \mathbf{1}$

Now we know Z is 1. Let us look at the units column. $Y + Z = O \quad \text{or} \quad Y + Z = 10 + O$ Let us call the carry from units column $c_1$ . So, $c_1$ can be 0 or 1.

Let us look at the tens column. $Y + c_1 = O \quad \text{or} \quad Y + c_1 = 10 + O$ Let us call the carry from tens column $c_2$ . We found $c_2 = Z$ . Since $Z = 1$ , we know $c_2 = 1$ . This means the sum in the tens column must be 10 or more. So, $Y + c_1 = 10 + O$ .

Now we have two equations. From units column: $Y + 1 = O$ (if $c_1=0$ ) or $Y + 1 = 10 + O$ (if $c_1=1$ ). From tens column: $Y + c_1 = 10 + O$ .

If $c_1 = 0$ : Then $Y + 1 = O$ . And $Y + 0 = 10 + O$ . So, $Y = 10 + O$ . Substitute $O = Y+1$ into $Y = 10+O$ . $Y = 10 + (Y + 1)$ $Y = 11 + Y$ $0 = 11$ This is not possible. So $c_1$ cannot be 0.

Therefore, $c_1$ must be 1. From units column: $Y + 1 = 10 + O$ . This simplifies to $Y - O = 9$ . Since Y and O are single digits, the only solution is $Y=9$ and $O=0$ .

$\text{Y} = \mathbf{9}$ $\text{O} = \mathbf{0}$

Let us check our values: Z=1, Y=9, O=0. All letters are distinct digits. YY is 99. Z is 1. ZOO is 100. $99 + 1 = 100$ This is correct.

$\boxed{\text{Y=9, Z=1, O=0}}$

Diagram 1

Step 2 — Solving (ii) B5 + 3D = ED5

Let us look at the units column. $5 + D = 5 \quad \text{or} \quad 5 + D = 10 + 5$ If $5 + D = 10 + 5$ , then $D = 10$ . This is not possible, as D must be a single digit. So, $5 + D = 5$ . This means D must be 0. The carry to the tens column is 0. Let us call it $c_1$ .

$\text{D} = \mathbf{0}$ $c_1 = \mathbf{0}$

Now we know D is 0 and $c_1$ is 0. Let us look at the tens column. $B + 3 + c_1 = D \quad \text{or} \quad B + 3 + c_1 = 10 + D$ Substitute $c_1=0$ and $D=0$ . $B + 3 + 0 = 0 \quad \text{or} \quad B + 3 + 0 = 10 + 0$ If $B + 3 = 0$ , then $B = -3$ . This is not possible. So, $B + 3 = 10$ . This means B must be 7. The carry to the hundreds column is 1. Let us call it $c_2$ .

$\text{B} = \mathbf{7}$ $c_2 = \mathbf{1}$

Now we know B is 7 and $c_2$ is 1. Let us look at the hundreds column. $c_2 = E$ Since $c_2 = 1$ , E must be 1.

$\text{E} = \mathbf{1}$

Let us check our values: B=7, D=0, E=1. All letters are distinct digits. B5 is 75. 3D is 30. ED5 is 105. $75 + 30 = 105$ This is correct.

$\boxed{\text{B=7, D=0, E=1}}$

Diagram 2

Step 3 — Solving (iii) KP + KP = PRR

Let us look at the hundreds column. The sum of KP and KP gives PRR. The carry to the hundreds column must be P. The maximum sum of $99 + 99$ is $198$ . So, the carry to the hundreds column can only be 1. This means P must be 1.

$\text{P} = \mathbf{1}$

Now we know P is 1. Let us look at the units column. $P + P = R \quad \text{or} \quad P + P = 10 + R$ Substitute P=1. $1 + 1 = R \quad \text{or} \quad 1 + 1 = 10 + R$ If $1 + 1 = R$ , then $R = 2$ . The carry to the tens column is 0. Let us call it $c_1$ . If $1 + 1 = 10 + R$ , then $2 = 10 + R$ . This means $R = -8$ , which is not possible. So, R must be 2.

$\text{R} = \mathbf{2}$ $c_1 = \mathbf{0}$

Now we know P=1, R=2, and $c_1=0$ . Let us look at the tens column. $K + K + c_1 = R \quad \text{or} \quad K + K + c_1 = 10 + R$ We know the carry to the hundreds column ( $c_2$ ) is P, which is 1. So, the sum in the tens column must be 10 or more. $K + K + c_1 = 10 + R$ Substitute $c_1=0$ and $R=2$ . $K + K + 0 = 10 + 2$ $2K = 12$ This means K must be 6.

$\text{K} = \mathbf{6}$

Let us check our values: K=6, P=1, R=2. All letters are distinct digits. KP is 61. PRR is 122. $61 + 61 = 122$ This is correct.

$\boxed{\text{K=6, P=1, R=2}}$

Diagram 3

Step 4 — Solving (iv) C1 + C = 1FF

Let us look at the hundreds column. The sum of C1 and C gives 1FF. The carry to the hundreds column must be 1. Let us call it $c_2$ .

$c_2 = \mathbf{1}$

Now we know $c_2$ is 1. Let us look at the tens column. $C + c_1 = F \quad \text{or} \quad C + c_1 = 10 + F$ Since $c_2=1$ , the sum in the tens column must be 10 or more. So, $C + c_1 = 10 + F$ .

Let us look at the units column. $1 + C = F \quad \text{or} \quad 1 + C = 10 + F$ Let us call the carry from units column $c_1$ .

If $c_1 = 0$ : Then from units column: $1 + C = F$ . From tens column: $C + 0 = 10 + F$ . So $C = 10 + F$ . Substitute $F = 1+C$ into $C = 10+F$ . $C = 10 + (1 + C)$ $C = 11 + C$ $0 = 11$ This is not possible. So $c_1$ cannot be 0.

Therefore, $c_1$ must be 1. From units column: $1 + C = 10 + F$ . This simplifies to $C - F = 9$ . Since C and F are single digits, the only solution is $C=9$ and $F=0$ .

$\text{C} = \mathbf{9}$ $\text{F} = \mathbf{0}$

Let us check our values: C=9, F=0. All letters are distinct digits. C1 is 91. C is 9. 1FF is 100. $91 + 9 = 100$ This is correct.

$\boxed{\text{C=9, F=0}}$

Diagram 4

Answer

(i) Y=9, Z=1, O=0 (ii) B=7, D=0, E=1 (iii) K=6, P=1, R=2 (iv) C=9, F=0

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Q41

Q. Find out what each letter stands for:

(i) $\begin{array}{r} \text{YY} \\ +\quad \text{Z} \\ \hline \text{ZOO} \end{array}$

(ii) $\begin{array}{r} \text{B5} \\ +\quad \text{3D} \\ \hline \text{ED5} \end{array}$

(iii) $\begin{array}{r} \text{KP} \\ +\quad \text{KP} \\ \hline \text{PRR} \end{array}$

(iv) $\begin{array}{r} \text{C1} \\ +\quad \text{C} \\ \hline \text{1FF} \end{array}$