A two-digit number has tens digit greater than the unit's digit. If the sum of its digits is equal to twice the difference, how many such numbers are possible?

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Solution

Let the two numbers be $a, b$

Given,
$a > b$

$(a + b) = 2 (a - b)$

$⟹ a = 3 b$

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

$b = 1, 2, 3$

$a = 3, 6, 9$

So, the possible numbers are $31, 62, 93$ .

Hence, $3$ such numbers are possible.

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A two-digit number has tens digit greater than the unit's digit. If the sum of its digits is equal to twice the difference, how many such numbers are possible?

Let the two numbers be a,bGiven, a>b(a+b)=2(a−b)⟹a=3bPut values of b such that a<10 (otherwise it would be a 3 digit number)b=1,2,3a=3,6,9So, the possible numbers are 31,62,93.Hence, 3 such numbers are possible.

Let the two numbers be $a, b$

Given,
$a > b$

$(a + b) = 2 (a - b)$

$⟹ a = 3 b$

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

$b = 1, 2, 3$

$a = 3, 6, 9$

So, the possible numbers are $31, 62, 93$ .

Hence, $3$ such numbers are possible.