Question

How many two-digit numbers with their tens digit
greater than their units digit, have the sum of their
digits equal to twice their difference?​

Answer 1

Let the number at tens place be A and units place be B,

Now according to the conditions,

A>B and A+B=2×(difference)

Since A>B, difference= A-B

A+B=2×(A-B)

A+B=2A- 2B

3B=A

So the numbers have A=3B

Put values of B such that A<10(otherwise it would be a 3 digit number)

So B=1,2,3 and A=3,6,9 respectively,

Hence the numbers are 31,62,93

hope this helps buddy

mark the brainliest

Answer 2

Answer:

answer is 31,62,9

thank you