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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, , Exercise 16.1, , Page No: 255, , Find the values of the letters in each of the following and give reasons for the steps, involved., 1., , Solution:, Say, A = 7 and we get,, 7+5 = 12, In which one’s place is 2., Therefore, A = 7, And putting 2 and carry over 1, we get, B=6, Hence A = 7 and B = 6, 2., , Solution:, If A = 5 and we get,, 8+5 = 13 in which ones place is 3., Therefore, A = 5 and carry over 1 then
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, B = 4 and C = 1, Hence, A = 5, B = 4 and C = 1, , 3., , Solution:, On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,, AxA = 6x6 = 36 in which ones place is 6., Therefore, A = 6, 4., , Solution:, Here, we observe that B = 5 so that 7+5 =12, Putting 2 at ones place and carry over 1 and A = 2, we get, 2+3+1 =6, Hence A = 2 and B =5
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, 5., , Solution:, Here on putting B = 0, we get 0x3 = 0., And A = 5, then 5x3 =15, A = 5 and C=1, Hence A = 5, B = 0 and C = 1, , 6., , Solution:, On putting B = 0, we get 0x5 = 0 and A = 5, then 5x5 =25, A = 5, C = 2, Hence A = 5, B = 0 and C =2, 7.
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, Solution:, Here product of B and 6 must be same as ones place digit as B., 6x1 = 6, 6x2 = 12, 6x3 = 18, 6x4 =24, On putting B = 4, we get the ones digit 4 and remaining two B’s value should be44., Therefore, for 6x7 = 42+2 =44, Hence A = 7 and B = 4, , 8., , Solution:, On putting B = 9, we get 9+1 = 10, Putting 0 at ones place and carry over 1, we get for A = 7, 7+1+1 =9, Hence, A = 7 and B = 9, , 9., , Solution:, On putting B = 7, we get 7+1 =8, Now A = 4, then 4+7 =11, Putting 1 at tens place and carry over 1, we get, 2+4+1 =7, Hence, A = 4 and B = 7, 10.
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, Solution:, Putting A = 8 and B = 1, we get, 8+1 =9, Now, again we add2 + 8 =10, Tens place digit is ‘0’ and carry over 1. Now 1+6+1 = 8 =A, Hence A = 8 and B =1
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, , Exercise 16.2, , Page No: 260, , 1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?, Solution:, Suppose 21y5 is a multiple of 9., Therefore, according to the divisibility rule of 9, the sum of all the digits should be a multiple of, 9., That is, 2+1+y+5 = 8+y, Therefore, 8+y is a factor 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., Therefore, the value of y should be 1 only i.e. 8+y = 8+1 = 9., 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:, Since, 31z5 is a multiple of 9., Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of, 9., 3+1+z+5 = 9+z, Therefore, 9+z is a multiple of 9, This is only possible when 9+z is any one of these numbers: 0, 9, 18, 27, and so on., This implies, 9+0 = 9 and 9+9 = 18, Hence 0 and 9 are two possible answers., 3. If 24x is a multiple of 3, where x is a digit, what is the value of x?, (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. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different, values.), Solution:, Let's say, 24x is a multiple of 3., Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3., 2+4+x = 6+x, So, 6+x is a multiple of 3, and 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 and so on., Since, x is a digit, the value of x will be either 0 or 3 or 6 or 9, and the sum of the digits can be, 6 or 9 or 12 or 15 respectively., Thus, x can have any of the four different values: 0 or 3 or 6 or 9., 4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
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NCERT Solutions for Class 8 Maths Chapter 16 Playing with, Numbers, Solution:, Since 31z5 is a multiple of 3., Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of, 3., That is, 3+1+z+5 = 9+z, Therefore, 9+z is a multiple of 3., This is possible when the value of 9+z is any of the values: 0, 3, 6, 9, 12, 15, and so on., At z = 0, 9+z = 9+0 = 9, At z = 3, 9+z = 9+3 = 12, At z = 6, 9+z = 9+6 = 15, At z = 9, 9+z = 9+9 = 18, The value of 9+z can be 9 or 12 or 15 or 18., Hence 0, 3, 6 or 9 are four possible answers for z.
