KSEEB Class 8 Maths Solutions For Chapter 1 Playing With Numbers Points To Remember
Numbers in general form:
- A number is said to be in a general form if it is expressed as the sum of the products of its digits with their respective place values.
Two-digit numbers:
- A two-digit number is written as, ab = 10a+b where a and b are whole numbers taking values from 0 to 9 such that , a ≠ 0
Example: 63 = 10 x 6 + 3
Three-digit numbers:
- A three digit number abc is written as abc = 100a +10b +c are whole numbers taking values from 0 to 9 such that a ≠0
Example: 765 = 100 x 7 +10 x 6 +5
Kseeb Solutions For Class 8 Maths Chapter 1 Playing With Numbers
Tests of Divisibility
Read and Learn More KSEEB Solutions for Class 8 Maths
Divisibility by 2 :
- A number is divisible by 2 when its one’s digit is0, 2, 4, 6 or 8.
Explanation: Given number
abc = 100a + 10b +c
100a and 10b are divisible by 2 because 100 and 10 are divisible by only when a= 0, 2, 4, 6 or 8.
| Class 10 Science | Class 11 Chemistry |
| Class 11 Chemistry | Transformation of Sentences |
| Class 8 Maths | Class 8 Science |
Divisibility by 3:
- A number is divisible by 3 when the sum of its digits is divisible by 3.
Example: Given number = 17658
Sum of digits = 1+7+6+5+8 =27 which is divisible by 3
∴ 17658 is divisible by 3
Divisibility by 4 :
- A number is divisible by 4 when the number formed by its last two digits is divisible by 4.
Example: 6216,548 etc.
Divisibility by 5:
- A number is divisible by 5 when its ones digit is 0 or 5 .
Example: 545,640 etc.
Divisibility by 6:
- A number is divisible by 6 when it is divisible by both 2 and 3 .
Example: 246,7230 etc.
Divisibility by 9:
- A number is divisible by 9 , when the sum of its digits is divisible by 9 .
Example: Consider a number 215847
Sum of digits = 2+1+5+8+4+7=27 which is divisible by 9.
∴ 215847 is divisible by 9 .
Divisibility by 10:
- A number is divisible by 10 when its one digit is 0 .
Example: 640,420 etc.
Letters for digits (puzzles)
- Many number puzzles involving different letters for different digits are solved using rules of number operations.
- Two rules :
- Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.
- The first digit of a number cannot be zero, thus, we write the number “sixty-three” as 63 and not as 063 , or 0063 .
- Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.
- The first digit of a number cannot be zero, thus, we write the number “sixty-three” as 63 and not as 063 , or 0063 .
Playing With Numbers Class 8 Kseeb Solutions Pdf
Playing With Numbers Solutions KSEEB Class 8 Maths Exercises 1.1
1 . Find the values of the letters in each of the following and give reasons for the steps involved.
\(\begin{array}{r}3 \mathrm{~A} \\
+25 \\
\hline \mathrm{B} 2 \\
\hline
\end{array}\)
Solution:
The addition of A and 5 is giving 2 i.e., a number whose ones digit is 2. This is possible only when digit A is 7. In that case, the addition of A(7) and 5 will give 12 and thus, I will be the carry for the next step. In the next step 1 + 3+2+ = 6
∴ The addition is as follows
\(\begin{array}{r}37 \\
+25 \\
\hline 62 \\
\hline
\end{array}\)
Clearly, B is 6
Hence, A and Bare 7 and 6 respectively.
2. \( \begin{array}{r}
4 \mathrm{~A} \\
+98 \\
\hline \mathrm{CB3} \\
\hline
\end{array} \)
Solution: The addition of A and 8 is giving 3 i.e, a number whose ones digit is 3 . This is possible only when digit A is 5 . In that case, the addition of A and 8 will give 13 and thus, 1 will be the carry for the next step. In the next step.
1+4+9 = 14
∴ The addition is as follows
\(\begin{array}{r}45 \\
+98 \\
\hline 143 \\
\hline
\end{array}\)
Clearly, B and C are 4 and 1 respectively. Hence, A,B and C are 5,4 and 1 respectively.
3. \(\begin{array}{r}
1 \mathrm{~A} \\
\times \mathrm{A} \\
\hline 9 \mathrm{~A} \\
\hline
\end{array}\)
Solution: The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A=1,5 or 6.
If A=1, then the multiplication will be 11 x 1 =1. However, here the tens digit is given as 9 .
∴ A=1 is not possible.
Thirdly, if A=5, then the multiplication will be 15 x 5 75, thus A = 5 is also not possible.
If we take A=6 , then 16 x 6 = 96
∴ A should be 6
The multiplication is as follows
\(\begin{array}{r}16 \\
\times 6 \\
\hline 96 \\
\hline
\end{array}\)
Hence, the value of A is 6.
Class 8 Maths Chapter 1 Kseeb Solutions Karnataka Board
4. \(\begin{array}{r}
\mathrm{AB} \\
+37 \\
\hline 6 \mathrm{~A} \\
\hline
\end{array}\)
Solution: The addition of A and 3 is giving 6 . There can be two cases.
- 1st step is not producing a carry
In that case, A comes to be 3 as 3 + 3 = 6 considering the first step in which the addition of B and 7 is giving A (i.e., 3), B should be a number such that the unit digit of this addition comes to be 3 , it is possible, only when B = 6. In this case,A = 6 + 7= 13. However, A is a single digit number, hence it is not possible. - 2nd step is producing a carry
In that case, A comes to be 2 as 1 + 2 + 3 = 6, considering the first step in which the addition of B and 7 is giving A (i.e., 2) B should be a number such that the units digit of this addition comes to be 2 . It is possible only when B = 5 and 5 + 7 = 12.
25 \\
+37 \\
\hline 62 \\
\hline
\end{array}\)
Hence, the values of A and B are 2 and 5 respectively.
5. \(\begin{array}{r}
\text { A B } \\
\times 3 \\
\hline \text { CAB } \\
\hline
\end{array}\)
Solution: The multiplication of 3 and B gives a number whose ones digit is B again. Hence B must be 0 or 5 .
Let B is 5.
Multiplication of 1st step = 3 x 5 = 15
1 will be a carry for the next step.
We have, 3 x A +1 = CA
This is not possible for any value of A. Hence, B must be 0 only. If B = 0, then there will be no carry for the next step we should obtain, 3 x A= CA . That is, the one’s digit of 3 x A should be A. This is possible When A=5 or 0.
However, A cannot be 0 as AB is a two digit number.
∴ A must be 5 only, the multiplication is as follows.
\(\begin{array}{r}50 \\
\times 3 \\
\hline 150 \\
\hline
\end{array}\)
Hence, the values of A, B and C are 5, 0 and 1 respectively.
Kseeb Solutions For Class 8 Maths Chapter 1 Important Questions
6. \(\begin{array}{r}
\mathrm{AB} \\
\times 5 \\
\hline \mathrm{CAB} \\
\hline
\end{array}\)
Solution: The multiplication of B and 5 is giving a number whose ones digit is B again this is possible when B=5 or B=0 only. In case of B=5 . The product, B x 5 = 5 x 5 = 25.5 will be a carry for the next step. We have,5 x A = 2 = CA, which is possible for A = 2 or 7
The multiplication is as follows
\(\begin{array}{r}
25 \\
\times 5 \\
\hline 125 \\
\hline
\end{array}\)
\(\begin{array}{r}
75 \\
\times 55 \\
\hline 375 \\
\hline
\end{array}\)
If B=0
B x 5 = B ⇒ 0 x 5 = 0
There will not be any carry in this step.
In the next step, 5 x A = CA
It can happen only when A = 5 orA = 0
However, A cannot be 0 as AB is a two digit number.
Hence, A can be 5 only, the multiplication is as follows.
\(\begin{array}{r}50 \\
\times 5 \\
\hline 250 \\
\hline
\end{array}\)
Hence, there are 3 possible values of A, B and C.
1) 5,0 , and 2 respectively
2) 2,5 and 1 respectively
3) 7,5 and 3 respectively
7. \(\begin{array}{r}
\text { AB } \\
\times 6 \\
\hline \text { B BB B } \\
\hline
\end{array}\)
Solution: The multiplication of 6 and B gives a number whose one’s digit is B again. It is possible only when B =0,2,4,6 or 8. If B =0, then the product will be 0.
∴This value of B is not possible.
If B=2, then B x 6 = 12 and 1 will be a carry for the next step.
6A +1=BB=22 ⇒ 6A = 21and hence, any integer value of A is not possible.
If B=6 , then B x 6 = 36 and 3 will be a carry for the next step.
6A+3 =B=66 ⇒ 6A=63 and hence, any integer value of is not possible.
If B=8 , then B x 6 =48 and 4 will be a carry for the next step.
6A +4=BB=88 ⇒ 6A = 84 and hence,A =14 however, is a single digit number.
∴This value of A is not possible.
If B=4, then B x 6 =24 and 2 will be a carry for the next step.
6A +2=BB=44
⇒ 6A = 42 and hence A = 7 The multiplication is as follows
\(\begin{array}{r}74 \\
\times 6 \\
\hline 444 \\
\hline
\end{array}\)
Hence, the values of A and B are 7 and 4 respectively.
8. \(\begin{array}{r}
\mathrm{A} 1 \\
+\underline{1 \mathrm{~B}} \\
\underline{\mathrm{BO}}
\end{array}\)
Solution: The addition of 1 and B is giving 0 . i.e., a number whose ones digits is 0 . This is possible only when digit B , is 9 . In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step. 1+A+1=B
Clearly, A is 7 as 1+7+1 =9 = B
∴The addition is as follows
\(\begin{array}{r}71 \\
+19 \\
\hline 90 \\
\hline
\end{array}\)
Hence, the values of A and B are 7 and 9 respectively.
9. \(\begin{array}{r}
2 \mathrm{AB} \\
+\mathrm{AB} 1 \\
\hline \mathrm{B} 18 \\
\hline
\end{array}\)
Solution: The addition of B and 1 is giving 8 i.e., a number, whose ones digit is 8 . This is possible only when digit B is 7 . In that case, the addition of B and 1 will give 8.
In the next step,
A+B=1
Clearly, A is 4 .
4+7=11 and 1 will be a carry for the next step. In the next step, 1+2+A=B
1+2+4=7
∴ the addition is as follows
\(\begin{array}{r}247 \\
+471 \\
\hline 718 \\
\hline
\end{array}\)
Hence the values of A and B are 4 and 7 respectively.
Karnataka Board Class 8 Maths Chapter 1 Textbook Solutions
10. \(\begin{array}{r}
12 \mathrm{~A} \\
+6 \mathrm{AB} \\
\hline \mathrm{A} 09 \\
\hline
\end{array}\)
Solution: The addition of A and B is giving 9 i.e., a number whose ones digits is 9 . The sum can be 9 only as the sum of two single digit numbers cannot be 19 .
∴ There will not be any carry in this step.
In the next step, 2+A=0
It is possible only when A=8
2+8=10 and 1 will be the carry for the next step 1+1+6=A
Clearly, A is 8 , we know that the addition of A and B is giving 9. As A is 8 .
is 1
∴The addition is as follows.
\(\begin{array}{r}128 \\
+681 \\
\hline 809 \\
\hline
\end{array}\)
Hence, the values of A and B are 8 and 1 respectively.
Playing With Numbers Problems In KSEEB Maths
KSEEB Class 8 Maths Chapter 1 Playing With Numbers Exercise 1.2
1. If 21y5 is a multiple of 9 , where y is a digit. What is the value of y ?
Solution: If a number is a multiple of 9 , then the sum of its digits will be divisible by 9 .
Sum of digits of 21y5=2+1+y+5= 8+y
Hence, 8+y should be a multiple of 9 .
This is possible when 8+y is any one of these numbers 0,9,18,27 and so on.
However, since y is a single digit number. This sum can be 9 only.
∴ y should be 1 only.
2. If 31z5 is a multiple of 9 , where z is a digit, what is the value of z ? You will find that there are two answers for the last problem. Why is this so?
Solution: If a number is a multiple of 9 , then the sum of its digits will be divisible by 9 .
Sum of digits of 31z5 = 3+1+z+5+9 = 9+z.
Hence, 9+z should be a multiple of 9 . This is possible when 9+z is any one of these numbers 0,9,18,27 and so on.
However, since is a single digit number this sum can be cither 9 or 18 .
∴ z should be either 0 or 9.
3. If 24x is a multiple of 3 , where x is a digit, what is the value of x ?
Solution: Since 24x is a multiple of 3 , the sum of its digits is a multiple of 3 , sum of digits of 24x = 2+4+x=6+x
Hence, 6 is a multiple of 3 .
This is possible when 6+x is any one of these numbers 0,3,6,9 and so on.
Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.
Thus can have its value as any of the four different values or 9 .
Kseeb Solutions For Class 8 Maths Playing With Numbers Exercises
4. If 31z5 is a multiple of 3 , where is a digit, what might be the values of z ?
Solution: Since 31z5 is a multiple of 3 , the sum of its digits will be a multiple of 3 . i.e., 3+1+z+5=9+z is a multiple of 3. This is possible when 9+z is any one of 0,3,6,9,12,15,18 and so on.
Since z is a single-digit number, the value of 9+z can only be 9 or 12 or 15 or 18 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.
Thus z can have its value as any one of the four different values 0,3,6 or 9 .
KSEEB Maths Class 8 Playing With Numbers Additional Problems
1. Without actual division find the remainder when 3,79,843 is divided by 3 .
Solution: We can find the remainder by dividing the sum of all the digits of the given number = 3+7+9+8+4+3=34. When 34 divided by 3 , we get 1 as remainder.
Hence, division 3,79,843 of by 3 leaves remainder of 1.
2. Find the values of letter A and B.
\(\begin{array}{r}2 \mathrm{AB} \\
+\mathrm{AB} 1 \\
\hline \mathrm{B} 18 \\
\hline
\end{array}\)
Solution: B+1=8 ⇒ B=8-1=7
A+B =1⇒ A+7=1 ( is ones digit)
A+7 can either 11 or 21 .
When A+7=11 ⇒ A=11-7=4
When A+7=21 ⇒ A=21-7=14 (not possible)
Thus A=4
∴ \(\begin{array}{r}
2 \text { A B } \\
+\text { A B 1 } \\
\hline \text { B 18 } \\
\hline
\end{array}\)
is \(\begin{array}{r}
247 \\
+471 \\
\hline 718 \\
\hline
\end{array}\)
∴ A is 4 and B is 7
Class 8 Maths Playing With Numbers Kseeb Solved Examples
3. If from a two digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible? Write all of them.
Solution: Let ab be any two digit number. Then the digit formed by reversing its digits is ba.
Now, \(\begin{aligned}
a b-b a & =(10 a+b)-(10 b+a) \\
& =(10 a-a)+(b-10 b) \\
& =9 a-9 b=9(a-b)
\end{aligned}\)
Further, since ab – ba is a perfect cube and is a multiple of 9.
∴ The possible value of a-b is 3
i.e., a = b+3
Here, b can take value from 0 to 6
For b = 0, a = 3 i.e., 3 0
For b = 1, a = 4, i.e. 4 1
For b = 2, a = 5, i.e., 5 2
For b = 3, a = 6, i.e., 6 3
Forb = 4, a = 7, i.e., 7 4
For b = 5, a= 8, i.e., 8 5
For b = 6, a= 9, i.e., 9 6
4. Find the value of the letters in each of the following :
1.\(\begin{array}{r}
\mathrm{PQ} \\
\times 6 \\
\hline \mathrm{QQQ} \\
\hline
\end{array}\)
2. \( \begin{array}{r}
2 \text { L M } \\
+\text { L M 1 } \\
\hline \text { M18 } \\
\hline
\end{array}\)
Solution: 1) We have, \( \begin{array}{r}
\mathrm{PQ} \\
\times 6 \\
\hline \mathrm{QQQ} \\
\hline
\end{array}\)
Here, in first column, we see that 6 x QP = Q
∴ The possible values of Q are 2,4,6 and 8
For Q=2, 6 X P +1 cannot be equal to 22 for any value of P.
So, Q=2 is not possible
For Q=4
⇒ 6 x p +2= 44
⇒ 6P = 44-2=42
⇒ P=2
Hence P =7 and Q = 4
2. \(\begin{array}{r}
2 \text { L M } \\
+\text { L M 1 } \\
\hline \text { M1 } 8 \\
\hline
\end{array}\)
In first column M+1=8
Clearly, M=7
In second column L+M=1
⇒ L+7=1
∴The value of L can be 4
In third column, 2+L+1=M
⇒ 2+4+1=7
⇒ 7=7, so the third column is satisfied for L=4, M=7
Hence, L=4 andM=7
5. If \(\begin{array}{r}
1 \mathrm{P} \\
\times \mathrm{P} \\
\hline \mathrm{Q} 6 \\
\hline
\end{array}\) where Q-P=3 , then find the values of P and Q
Solution: We have \(\begin{array}{r}
1 \mathrm{P} \\
\times \mathrm{P} \\
\hline \mathrm{Q} 6 \\
\hline
\end{array}\)and Q-P=3
Here, P x P is 6 , so the value of P is either 4 or 6 . But, if P=4,Q=5, which does not satisfy the relation Q-P=3
Hence,P=6 and then Q=9 .
6. If A3+8B=150, then find the value of A+B.
Solution: We have A3+8B=150 .
Here 3+B=0, So 3+B is a two-digit number whose unit’s digit is zero.
∴ 3+B=10 ⇒ B=7
Now, considering ten’s column, A+8+1=15
⇒ A+9=15
⇒ A=6
Hence A+B=6+7=13
A3+8B=150
i.e., 63+87=150
150=150
Kseeb Class 8 Playing With Numbers Solutions Step By Step
7. If 56 x 32y is divisible by 18 , find the least value of y .
Solution: It is given that, the number 56 x 32y is divisible by 18 , then it is also divisible by each factor of 18. Thus, it is divisible by 2 as well as 3 . Now, the number is divisible by 2 its unit digit must be an even number that is 0,2,4,6.
∴The least value of y is 0
Again, the number is divisible by 3 also, sum of its digits is a multiple of 3 , ie., 5+6+x+3+2+y is a multiple of 3.
⇒ 16+x+y+=0,3,6,9,………
⇒ 16+x=18
⇒ x=2, which is the least value of x.
∴ x=2 and y=0
Practice Questions For Playing With Numbers KSEEB Maths
8. Without performing actual division, find the remainder when 430346 is divided by 9.
Solution: Here sum of digits 4+3+0+3+4+6 is 20. When 20 is divide by 9 then the remainder will be 2 .
9. In a two digit number the digit in the one’s place is three times the digit in the ten’s place and the sum of the digits is equal to 8 . What is the number?
Solution: Let ten’s digit be ‘ a ‘ then unit digit be ‘3a ‘
Number= 10 x a +3a
=10a+3a=13a
According to the question
a+3a=8
4a=8
a=2
∴ Number = 13 x 2 =26
10. If 24x is a multiple of 3 , where x is a digit, what is the value of x?
Solution: Since 24x is a multiple of 3 its sum of digits 6+x is a multiple of 3 . So 6+x is one of these numbers;0,3,6,9,12,15,18, ………….
But since x is a digit, it can only be that 6+x=6 or 9 or 12 or 15.
∴ x=0 or 3 or 6 or 9.
Thus, x can have any of four different values.
11. Find the values of the letter?
\(\begin{gathered}
\mathrm{AB} \\
\times 5 \\
\hline \mathrm{CAB}
\end{gathered}\)
Solution: This means that 5 x B is a number whose units digit is B . Clearly, B can take value 5 .
Taking B=5, we have
\(\Rightarrow \overline{\mathrm{A} 5} \times 5=\overline{\mathrm{CA} 5}\) \(\Rightarrow(10 A+5) \times 5=100 \times C+A \times 10+5 \times 1\) \(\Rightarrow 10 A+5=\frac{1}{5}(100 C+10 A+5)\) \(\Rightarrow 10 A+5=20 C+2 A+1\) \(\Rightarrow 8 A+4=20 C \Rightarrow A(2 A+1)=20 C\)\(\Rightarrow 2 A+1=5 C\)
⇒ 2A + is an odd multiple of 5, hence 2A+1 is odd [∴ o< A≤ 9]
∴ 2A+1=5 or 2A+1=15
A=2, A=7
By putting A=2 in (i) we get C = 1
⇒ A=2, B=5 and C=1
and by putting A=7 in (i) we get C=3
∴ A=7,B=5 and C=3
Karnataka Board Class 8 Maths Solutions For Playing With Numbers
12.The ratio of tens digit to the units digit of a two digit number is 2:3. If 27 is added to the number, the digits interchange their places, find the number.
Solution: Let the unit digit of the required number be X and the tens digit of the required number be Y
then the required number is and number obtained by reversing the digits is also \(\frac{y}{x}=\frac{2}{3}\)
\(y=\frac{2}{3} x\) ……..(i)
According to the question
10 y+x+27=10 x+y ………(ii)
Putting \(y=\frac{2}{3} x\) from equation (i) in equation (ii)
\(10 \times \frac{2}{3} x+x+27=10 x+\frac{2}{3} x\) \(\frac{20}{3} x+x+27=\frac{30 x+2 x}{3}\) \(\frac{20 x+3 x}{3}+27=\frac{32 x}{3}\) \(23 x+27 \times 3=32 x\)32 x-23 x=81
9 x=81
x=81 / 9
\(y=2 / 3 x=\frac{2}{3} \times 9=6\)Thus the required number is 10 x 6+9=69
Playing With Numbers Questions And Answers KSEEB Maths
13. Find the value of k where 31k2 is divisible by 6 .
Solution: 31k2 is divisible by 6 , then, it is also divisible by 2 and 3 both.
Now, 31k2 is divisible by 3 , sum of its digits is a multiple of 3.
⇒ k+6=0,3,6,9,12……..
⇒ k=0 or 3,6,9.
14. Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9.
1. 59
2. 203
Solution: 1. Given number = 59
Number obtained by reversing the digits = 95
Difference = 95-59=36÷9=4
Hence, the required number is 9.
2) Given number = 203.
Number obtained by reversing the digits = 302
Difference = 302-203=99÷9=11
Hence, the required number is 9.
15. If 1AB + CCA = 697 and there is no carry over in addition, find the value of A+B+C.
Solution: We have 1AB
\(\frac{+\mathrm{CAA}}{697}\)Since there is no carry over – in addition
⇒ 1+C=6
C=5
⇒ A+C=9
A+5=9
⇒ A=4
B+A=7
⇒ B=3
Hence, A+B+C=4+3+5=12
16. Observe the following patterns
1×9-1=8
21×9-1=188
321×9-1=2888
4321×9-1=38888
Find the value of 87654321×9-1
Solution: From the pattern, we observe that there are as many eights in the result as the first digit from the right which is to be multiplied by 9 and reduced by 1
87654321×9-1 = 788888888.
17. In a 3 – digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digits Also, the sum of its digits is 18 . Find the number.
Solution: Let the tens digit be x
then, the hundreds digit = 2x and the unit digit =3x
∴ 2x+x+3x=18 ⇒ 6x=18 ⇒ x=3
∴ hundreds digit=(2×3)=6,
tens digit = 3 and
units digit=(3×3)=9
Hence, the required number
=(100×6+10×3+9)=639
Kseeb Solutions For Class 8 Maths Pdf
18. Find all possible values of x for which the 4-digit number 754x si divisible by 3 . Also, find each such number.
Solution: It is given that the number 754x is divisible by 3.
∴ Sum of its digits = (7+5+4+x) must be.
This happens when x=2 or 5 or 8 divisible by 3 . Since x a digit, it cannot be more than 9.
∴ x=2 or x=5 or x=8 or are only required values of . Hence, all required numbers are 7542,7545,7548
19. If a,b, c are three digits of a three digit number, prove that abc+ cab+ bca is a multiple of 37.
Solution: We have abc + cab + bca
abc=100a+10b+c
cab=100c+10a+b
bca=100b+10c+a
adding abc+cab+bca = 111a+111b+111c
=111(a+b+c) = 3 x 3(a+b+c)
which is a multiple of 37.
Hence proved.
20. The product of two 2 – digit numbers is 1431 . The product of their tens digits is 10 and the product of their units digits is 21 . Find the numbers.
Solution: Let the required two 2 digit numbers be 10a+b and 10p+q as per the condition.
We have a x p=10 and b x q=21
a =2 and p=5 or a=5 and p=2
111ly b x q=21, b=3 and q = 7 or b =7 and q =3
10p+q=57 or 10p+q =53 and
10a+b =23 or 10a+b =27
Since the units digit of product 1431 is 1 numbers are 57 & 23 or 53 & 27
Now & which is given. Hence, the required numbers are 53 & 27.