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:
If ab is the two digit number in its casual form and its reverse is ba. Writing the both numbers in generalised form we have:
10a + b --- (1)
10b + a --- (2)
Subtracting (2) from (1) we have
9a - 9b
So if 9(a - b) is a perfect cube means:
Now 9 = 3²
9(a - b) = 3² × (a - b)
So if (a - b) is 3 then the number 9(a - b) becomes a perfect cube i.e. 27
Therefore the values of a and b which make a-b = 3
The numbers are 30, 41, 52, 63, 74, 85, 96,
✦ Try This: If from a three-digit number, we subtract the number formed by reversing its digits then the result so obtained is divisible by ______ and ______.
The three digits number ‘abc’ can be written in its generalised form as 100a + 10b + c
The reverse of the three digit number ‘cba’ can be written in its generalized form as 100c + 10b + a
100a + 10b + c - (100c + 10b + a) = 99a - 99c = 99(a - c)
The resultant number is divisible by 9 and 11.
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16
NCERT Exemplar Class 8 Maths Chapter 13 Problem 69
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.
Summary:
30, 41, 52, 63, 74, 85, 96 are perfect cubes which are formed by subtracting the number formed by reversing its digits from the two digit number.
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