Answer:
Step-by-step explanation:
From the information given,
Let the two digits that Chen writes down to be p and q
where,
q will be in the unit place and p will be in the tens place.
So, the value of the digit = 10p + q
So, if he swapped the number, he will have 10q + p
However,
the new number determined is thrice more than the 1/3 pf the original number.
i.e
the 1/3rd of the original number = [tex]\dfrac{1}{3}(10p+q)[/tex]
=[tex]\dfrac{10p+q}{3}[/tex]
thrice more than that will be = [tex]\dfrac{10p+q}{3} + 3[/tex]
∴
[tex]\dfrac{10p+q}{3} + 3 = 10q+p[/tex]
multiply through by 3, we have :
10p+q + 9 = 30q + 3p
collecting the like terms, we have:
10p - 3p +q -30q +9 = 0
7p - 29q +9 = 0
7p - 29q = -9
Therefore, p,q lie between 0 and 9
Therefore, the possibility of the original number is p = 7 and q = 2
