Find the values of the letters in each of the following and give reasons for the steps involved.
1.
\( \begin{array}
\hline
\;\;\;3 & A\\
+ 2 & 5\\
\hline
\;\;B & 2\\
\hline
\end{array}
\)
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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.
\( \begin{array}
\hline
\;\;\;4 & A\\
+ 9 & 8\\
\hline
C\;B & 3\\
\hline
\end{array}
\)
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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.
\(\;\; \begin{array}
\hline
\;\;\;1 & A\\
\;\;×& A\\
\hline
\;\;9 & A\\
\hline
\end{array}
\)
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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.
\( \;\;\begin{array}
\hline
\;\;\;A & B\\
+\;3& 7\\
\hline
\;\;6 & A\\
\hline
\end{array}
\)
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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. \( \begin{array}
\hline
\;\;\;A & B\\
\times& 3\\
\hline
C \;\;A & B\\
\hline
\end{array}
\)
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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.
\(\;\;\; \begin{array}
\hline
\;\;\;A & B\\
\times& 5\\
\hline
C \;\;A & B\\
\hline
\end{array}
\)
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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.
\(\;\; \begin{array}
\hline
\;\;\;A & B\\
\times& 6\\
\hline
B\;\;B & B\\
\hline
\end{array}
\)
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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.
\( \;\;\begin{array}
\hline
\;\;\;A & 1\\
+ 1 & B\\
\hline
\;\;B & 0\\
\hline
\end{array}
\)
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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.
\( \;\;\begin{array}
\hline
\;\;\;2 & A &B\\
+ A & B&1\\
\hline
\;\;B&1 & 8\\
\hline
\end{array}
\)
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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.
\(\;\; \begin{array}
\hline
\;\;\;1 & 2 &A\\
+ 6 & A&B\\
\hline
\;\;A&0 & 9\\
\hline
\end{array}
\)
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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.