NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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 :
And we can clearly see that value of B=6.
Therefore we got the values as A=7 and B=6
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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 :
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
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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:
We can see clearly that it is not similar with the given multiplication so A
For A=5:
We can see clearly that it is not similar with the given multiplication so A
For A=6:
We can see clearly that it is similar with the given multiplication so A=6.
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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 :
Hence, the assumption is correct and values are A=2 and B=5
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
We can observe that we have equations:
(i)B
Solving it we can have B=0 or 5
5 is not taken because, for satisfying the condition A
(ii)A
we can satisfy this condition only if A=5 i.e., 5
So we have A=5, B=0 and C=1
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
We can observe that we have equations:
(i)B
Solving it we can have B= 0 or 5
5 is removed because for satisfying the condition A
(ii)A
we can satisfy this condition only if A=5 i.e., 5
So we have A=5, B=0 and C=2
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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
We know this is only possible if A=7
Hence, we have got A=7 and B=4
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
We have condition as:
B+1=8.So, we got B=7
Also we have A+B=1
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
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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
Hence, the required value of y = 1.
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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:
For k = 2:
We have, 3 + 1 + z + 5 = 18
k = 3 is not possible because 3 + 1 + z + 5 = 27
Hence the required value of z is 0 or 9
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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
Thus, x can have any of the four different values, namely, 0,3,6 or 9.
NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers
Answer :
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
Hence, z can have any of the four different values, which are 0, 3, 6 or 9.