The three-digit number whose hundred's digit is the cube of its unit's digit and the ten's digit is the sum of unit's and hundred's digit is ___.

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Solution

Any three digit number can not have its hundred's digit more than 9. If the unit's digit is 'x', then the hundred's digit is $x^{3}$ .
$\Rightarrow x^{3} < 9$
$\Rightarrow x = 1 o r 2$
When x=1, the ten's digit is $1 + 1 = 2$
When x=2, the ten's digit is $8 + 2 = 10$ which is greater than 9 and hence $x \neq 2$ .
Therefore, the number is 121.

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The three-digit number whose hundred's digit is the cube of its unit's digit and the ten's digit is the sum of unit's and hundred's digit is ___.

Any three digit number can not have its hundred's digit more than 9. If the unit's digit is 'x', then the hundred's digit is x3. ⇒x3<9 ⇒x=1 or 2 When x=1, the ten's digit is 1+1=2 When x=2, the ten's digit is 8+2=10 which is greater than 9 and hence x≠2. Therefore, the number is 121.