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# 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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