**
Find the values of the letters in each of the following and give reasons for the steps involved.
1.
\( \begin{array}
\hline
\;\;\;3 & A\\
+ 2 & 5\\
\hline
\;\;B & 2\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that A+5 gives 2 as the unit's digit to the sum.So it's only possible if A=7.

Now, the sum becomes :

\( \begin{array}
\hline
\;\;\;3 & 7\\
+ 2 & 5\\
\hline
\;\;6 & 2\\
\hline
\end{array}
\)

And we can clearly see that value of B=6.

Therefore we got the values as A=7 and B=6

**
2.
\( \begin{array}
\hline
\;\;\;4 & A\\
+ 9 & 8\\
\hline
C\;B & 3\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that A+8 gives 3 as the unit's digit to the sum.So it's only possible if A=5.

Now, the sum becomes :

\( \begin{array}
\hline
\;\;\;4 & 5\\
+ 9 & 8\\
\hline
1\;\;4 & 3\\
\hline
\end{array}
\)

By this, we can clearly see that value of B and C as 4 and 1 respectively.

Therefore we got the values as A=5, B=4 and C=1

**
3.\(\;\; \begin{array}
\hline
\;\;\;1 & A\\
\;\;×& A\\
\hline
\;\;9 & A\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that A×A gives A as the unit's digit to the multiplication, that means it's only possible if A=1, 6 and 5.

Checking for all the case:

For A=1: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 1\\
\;\;×& 1\\
\hline
\;\;1 & 1\\
\hline
\end{array}
\)

We can see clearly that it is not similar with the given multiplication so A\(\neq\)1.

For A=5: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 5\\
\;\;×& 5\\
\hline
\;\;7 & 5\\
\hline
\end{array}
\)

We can see clearly that it is not similar with the given multiplication so A\(\neq\)5.

For A=6: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 6\\
\;\;×& 6\\
\hline
\;\;9 & 6\\
\hline
\end{array}
\)

We can see clearly that it is similar with the given multiplication so A=6.

**
4. \( \;\;\begin{array}
\hline
\;\;\;A & B\\
+\;3& 7\\
\hline
\;\;6 & A\\
\hline
\end{array}
\)
**

Open in new tab *link*

We have here two conditions :

(i) B+7=A

(ii)A+3=6

By solving the second equation we get A=6-3=3

Also if we keep A=3 directly in eq. (i) we get B=3-7=negative value. So it can be said that a carry was taken so let the carry be 1.Therefore, now A=2+1(carry) and let us check if eq. (i) satisfies it or not.

We have B+7=12[∵ 1(carry)2(value of A)]

So we get B=12-7=5.

So the sum becomes :

\( \begin{array}
\hline
\;\;\;2 & 5\\
+\;3& 7\\
\hline
\;\;6 & 2\\
\hline
\end{array}
\)

Hence, the assumption is correct and values are A=2 and B=5

**
5. \( \begin{array}
\hline
\;\;\;A & B\\
\times& 3\\
\hline
C \;\;A & B\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that we have equations:

(i)B\(\times\)3=B

Solving it we can have B=0 or 5

5 is not taken because, for satisfying the condition A\(\times\)3=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.

\(∴B=0\)

(ii)A\(\times\)3=C A

we can satisfy this condition only if A=5 i.e., 5\(\times\)3=15.

So we have A=5, B=0 and C=1

**
6.\(\;\;\; \begin{array}
\hline
\;\;\;A & B\\
\times& 5\\
\hline
C \;\;A & B\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that we have equations:

(i)B\(\times\)5=B

Solving it we can have B= 0 or 5

5 is removed because for satisfying the condition A\(\times\)5=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.

\(∴B=0\)

(ii)A\(\times\)5=C A

we can satisfy this condition only if A=5 i.e., 5\(\times\)5=25.

So we have A=5, B=0 and C=2

**
7. \(\;\; \begin{array}
\hline
\;\;\;A & B\\
\times& 6\\
\hline
B\;\;B & B\\
\hline
\end{array}
\)
**

Open in new tab *link*

We can observe that: Possible values of BBB are 111, 222, 333, etc.

Let us divide these numbers by 6 and check the nummber which is completely divisible.

111 ÷ 6 =18, remainder 3 so, 111 is rejected.

222 ÷ 6 = 37, remainder 0 but the quotient 37 is not of the form A 2. So,222 is rejected.

333 ÷ 6 = 55, remainder 3 so, 333 is rejected.
444 ÷ 6 = 74, remainder 0 and also the quotient 74 is of the form A 4.

So, we can say the number B=4 is satisfying.Let's check and find value of A

\( \begin{array}
\hline
\;\;\;A & 4\\
\times& 6\\
\hline
4\;\;4 & 4\\
\hline
\end{array}
\)

We know this is only possible if A=7

Hence, we have got A=7 and B=4

**
8.\( \;\;\begin{array}
\hline
\;\;\;A & 1\\
+ 1 & B\\
\hline
\;\;B & 0\\
\hline
\end{array}
\)
**

Open in new tab *link*

We have condition as:

B+1=0 i.e, B must be 9 to get 0 at unit's digit of the addition.So, we got B=9

Also we have A+1+1(carry)=B

\(\Rightarrow\) A+2=9

\(\Rightarrow\) A=7

Hence, we have A=7 and B=9

**
9.\( \;\;\begin{array}
\hline
\;\;\;2 & A &B\\
+ A & B&1\\
\hline
\;\;B&1 & 8\\
\hline
\end{array}
\)
**

Open in new tab *link*

We have condition as:

B+1=8.So, we got B=7

Also we have A+B=1

\(\Rightarrow\) A+7=11

A must be 4 to get unit's digit of A+7 as 1.So, A=4

And the last condition we have 2+A+1(carry)=B

Checking the values of A and B if it satisfies or not:

We have A=4 and B=7.Substituting, we have:

LHS: 2+A+1=2+4+1=7

RHS: B=7

Hence the values are verified.So we have A=4 and B=7

**
10.\(\;\; \begin{array}
\hline
\;\;\;1 & 2 &A\\
+ 6 & A&B\\
\hline
\;\;A&0 & 9\\
\hline
\end{array}
\)
**

Open in new tab *link*

We have condition as:

(i)A+B=9.

(ii)2+A=10.

(iii)1+6+1(carry)=A

We can say by equation (iii) and equation (ii) that A=8

Also, we have A+B=9. Sustituting A=8, we get B=1

Hence, we have A=8 and B=1

For a number is divisible by 9 we know that the sum of its digits should also divisible by 9.

Here, Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y

So we should have (8 + y) ÷ 9 = 1

\(\Rightarrow 8 + y = 9\)

\(\Rightarrow y = 9 – 8 = 1\)

Hence, the required value of y = 1.

We know that a number is a multiple of 9 if the sum of its digits is also divisible by 9.

Here, Sum of the digits of 31z5 = 3 + 1 + z + 5

So we have , 3 + 1 + z + 5 = 9k where k is an integer.

Checking for different values of k

For k = 1:

\(\Rightarrow\) 3 + 1 + z + 5 = 9

\(\Rightarrow\) z = 9 – 9 = 0

For k = 2:

We have, 3 + 1 + z + 5 = 18

\(\Rightarrow\) z = 18 – 9 = 9

k = 3 is not possible because 3 + 1 + z + 5 = 27

\(\Rightarrow\) z = 27 – 9 = 18 which is not a digit.

Hence the required value of z is 0 or 9

We are given that 24 x is a multiple of 3. So, the sum of its digits 2 + 4 + x = (6 + x) should be a multiple of 3.

Therefore, (6 + x) is one of the numbers 0, 3, 6, 9, 12, 15, 18…so on but as per given x can only be a digit. Therefore, (6 + x) must be equal to 6 or 9 or 12 or 15.

i.e., 6 + x = 6 or 9 or 12 or 15

\(\Rightarrow\) x = 0 or 3 or 6 or 9

Thus, x can have any of the four different values, namely, 0,3,6 or 9.

We are given that 31z5 is a multiple of 3. So, the sum of its digits 3 + 1 + z + 5 = (9 + z) should be multiple of 3.

Therefore, (9 + z) is one of the numbers 0, 3, 6, 9, 12, 15, 18, …so on but given that z is a digit. Therefore (9 + z) must be equal to 9 or 12 or 15 or 18.

So we have, 9 + z = 9 or 12 or 15 or 18

\(\Rightarrow\) z = 0 or 3 or 6 or 9

Hence, z can have any of the four different values, which are 0, 3, 6 or 9.