Asked by Suredra
A number of two digit exceed 4 times the sum of the digit by 6 and it is increased by 9 on reversing the digit then find the number.
Answers
Answered by
bobpursley
10T+Ones=4(T+Ones)-6
and
10*ones +T=9+10Tens+Ones
let T be tens digit, B be ones digit.
10T+B=4T+4B-6
10B+T=10T+B+9
or
6T-3B=6
-9T+9B=9
or
T= (54+27)/(54-27)=81/27 =3
B= (54+54)/(27)=4
Number then is 34
check
A number of two digit exceed 4 times the sum of the digit by 6
34-6?=4(7)
28=28 checks.
and
10*ones +T=9+10Tens+Ones
let T be tens digit, B be ones digit.
10T+B=4T+4B-6
10B+T=10T+B+9
or
6T-3B=6
-9T+9B=9
or
T= (54+27)/(54-27)=81/27 =3
B= (54+54)/(27)=4
Number then is 34
check
A number of two digit exceed 4 times the sum of the digit by 6
34-6?=4(7)
28=28 checks.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.