Question

Chen writes down a two digit number. He finds out that if he swapped the numbers, it would be three more than the 1/3 of the original number. What was the ORIGINAL number?

Answer 1

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 = $\dfrac{1}{3}(10p+q)$

=$\dfrac{10p+q}{3}$

thrice more than that will be = $\dfrac{10p+q}{3} + 3$

∴

$\dfrac{10p+q}{3} + 3 = 10q+p$

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