Playing with numbers using any of the digits from 0 to 9 is discussed here.
A number is said to be in a generalized form if it is expressed as the sum of the product of its digits with their respective place values.
Thus, a two-digit number having a and b as its digits at the tens and the ones places respectively is written in the generalized form as 10a + b, i.e., in general, a two-digit number can be written as 10a + b, where ‘a’ can be any of the digits from 1 to 9 and ‘b’ can be any of the digits from 0 to 9.
Similarly, a three-digit number can be written in the generalized form as 100a + 10b + c, where ‘a’ can be any one of the digits from 1 to 9 while ‘b’ and ‘c’ can be any of the digits from 0 to 9.
For example:
The generalized forms of a few numbers are given below:
56 = 10 × 5 + 6;
37 = 10 × 3 + 7;
90 = 10 × 9 + 0;
129 = 100 × 1 + 2 × 10 + 9;
206 = 100 × 2 + 10 × 0 + 6;
700 = 100 × 7 + 10 × 0 + 0.
1. What is the original number, if the sum of the digits of a two-digit number is seven. By interchanging the digits is twenty seven more than the original number?
Solution:
Let the original number be 10a + b.
Then, ‘a’ is the tens digit and ‘b’ is the units digit.
Since the sum of the digits is 7,
Therefore a + b = 7,
i.e., b = 7 - a..
So, the original number is 10a + (7 - a).
Therefore, the number obtained by interchanging the digits is
10(7 - a) + a,
and so we have {10(7 - a) + a} — {10a + (7 - a)} = 27.
Solving this equation, we get
a = 2.
And so, b = 7 - 2
= 7 - 2
= 5.
Hence, the original number is 10a + b = 20 + 5 = 25.
2. In a two-digit number, the digit in the units place is four times the digit in the tens place and sum of the digits is equal to 10. What is the number?
Solution:
Let the original number be 10a + b
Then, b = 4a and a + b = 10.
We put b = 4a in a + b = 10 so that a + 4a = 10,
i.e., 5a = 10,
i.e., a = 2.
Therefore, a = 2
and b = 4a, [where we know a = 2]
b = 4 × 2
b = 8
Hence, the number is 10a + b = 20 + 8 = 28.
● Fun with Numbers
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8th Grade Math Practice
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