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Question Number 29326 by abdus salam last updated on 07/Feb/18
The unit digit of the square of a number  and the units digit of the cube of the   number are equal to the unit digit of  the number. How many values are   possible for the units digits of such  numbers?
$$\mathrm{The}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{number}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{values}\:\mathrm{are}\: \\ $$$$\mathrm{possible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{units}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{such} \\ $$$$\mathrm{numbers}? \\ $$
Answered by mrW2 last updated on 07/Feb/18
let a be the unit digit of such numbers.  from condition 1:  a^2 =10b+a  ⇒(a−1)a=10b  i.e. the last digit of the product from  (a−1) and a must be zero.  ⇒a=0,1,5,6    from condition 2:  a^3 =10b+a  ⇒(a−1)a(a+1)=10b  i.e. the last digit of the product from  (a−1),a and (a+1) must be zero.  ⇒a=0,1,4,5,6,9    ⇒to fulfill both conditions,  a=0,1,5,6
$${let}\:{a}\:{be}\:{the}\:{unit}\:{digit}\:{of}\:{such}\:{numbers}. \\ $$$${from}\:{condition}\:\mathrm{1}: \\ $$$${a}^{\mathrm{2}} =\mathrm{10}{b}+{a} \\ $$$$\Rightarrow\left({a}−\mathrm{1}\right){a}=\mathrm{10}{b} \\ $$$${i}.{e}.\:{the}\:{last}\:{digit}\:{of}\:{the}\:{product}\:{from} \\ $$$$\left({a}−\mathrm{1}\right)\:{and}\:{a}\:{must}\:{be}\:{zero}. \\ $$$$\Rightarrow{a}=\mathrm{0},\mathrm{1},\mathrm{5},\mathrm{6} \\ $$$$ \\ $$$${from}\:{condition}\:\mathrm{2}: \\ $$$${a}^{\mathrm{3}} =\mathrm{10}{b}+{a} \\ $$$$\Rightarrow\left({a}−\mathrm{1}\right){a}\left({a}+\mathrm{1}\right)=\mathrm{10}{b} \\ $$$${i}.{e}.\:{the}\:{last}\:{digit}\:{of}\:{the}\:{product}\:{from} \\ $$$$\left({a}−\mathrm{1}\right),{a}\:{and}\:\left({a}+\mathrm{1}\right)\:{must}\:{be}\:{zero}. \\ $$$$\Rightarrow{a}=\mathrm{0},\mathrm{1},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{9} \\ $$$$ \\ $$$$\Rightarrow{to}\:{fulfill}\:{both}\:{conditions}, \\ $$$${a}=\mathrm{0},\mathrm{1},\mathrm{5},\mathrm{6} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Feb/18
3x(3113n7 5!r!
Commented by mrW2 last updated on 07/Feb/18
Now I understand. Thank you sir!
Commented by Rasheed.Sindhi last updated on 07/Feb/18
5up3r8!!!
Commented by NECx last updated on 08/Feb/18
wow
$${wow} \\ $$

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