Chapter – 16: Playing with Numbers
Exercise 16.1
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
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 = 6×6 = 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
5.
Solution:
Here on putting B = 0, we get 0x3 = 0.
And A = 5, then 5×3 =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 5×5 =25
A = 5, C = 2
Hence A = 5, B = 0 and C =2
7.
Solution:
Here product of B and 6 must be same as ones place digit as B.
6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 =24
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be44.
Therefore, for 6×7 = 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.
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
Exercise 16.2
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?
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.
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