# In a two-digit number, digit at the ten’s place is twice the digit at unit’s place.

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In a two-digit number, digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.

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Let the digit at unit’s place be x.

The digit at the ten’s place is twice the digit at unit’s place.

∴ The digit at ten’s place = 2x

 Digit in units place Digit in tens place Number Original Number x 2x (2x × 10) + x= 20x + x= 21x NewNumber 2x x (x × 10) + 2x= 10x + 2x= 12x

Since, the sum of the original number and the new number is 66.

∴ 21x + 12x = 66

∴ 33x = 66

∴ x = $\cfrac{66}{33}$

∴ x = 2

∴ Original number = 21x = 21 × 2 = 42

∴ the original number is 42.