The sum of the digits of a two-digit number is 8. If 16 is added to the original number, the result is 3 times the original number with its digits reversed. Find the original number. I got: {t+u=8 10t+u=3(10u+t)+16 But I don't know how to solve it? And is it right?
62
how did you solve it
take this number as ab, so a+b=8 and (ab)+16=3(ba) which means 10a+b+16 =3(10b+a) If you solve the equation, you get 29b-7a=16 adn you know that a+b=8, if you solve them both, you get 36b=72 so b=2 and a=6 ab=62
wait what?
how is it (ab)+16=3(ba)?
The mistake you make is this: You should think that if you add 16 to the number, you get 3 times of this so 16 should be on the leftside
shouldn't it be 3(ab)+16
got it_?
no
how is (ab)+16=3(ba)?
The problem says " If 16 is added to the original number, the result is 3 times the original number with its digits reversed" which means if you add 16 to (ab), you get 3 times of(ba)
oh ok
i don't know you got 62
is it wrong?
when i solved 10a+b+16 =3(10b+a) I got 10a+16=29b+3a
no wait i got 7a+16=29b and i don't know what to do from there
10a+b+16=30b+3a so b+16=30b-7a and 29b-7a=16
shouldn't u subtract 3t though
ok now you geot the true equation 7a+16=29b... you know that a+b=8, for it is in the question.Now write these equations in two lines and solve them both
ok forget a and b..let's do it your way.. let the number be (ut).. you know that u+t=8 and 10u+t+16=3(10t+u)
10u+t+16=30t+3u so 7u-29t=-16 so 29t-7u=16
yea..
29t-7u=16 and u+t=8 so u=8-t. if you put 8-t in the first equation; 29t-7(8-t)=16 so 29t-56+7t=16 and t=2 here so u=6
so if t=2 and u=6, how do you know to put it 62? like how do you know to not put it as 26?
because in the beginning you named the number as (ut) which means u=6 and t=2 and ut is 62.. we solved it according to this information we put 10u instead of u so it is 62
o ok thanks so much
ur welcome:)
Join our real-time social learning platform and learn together with your friends!