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.