# A two digit no. is such that the product of the digits is 14. When 45 is added to the no., then the digits interchange their places. Find the no.

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To answer this, first think of all the numbers from 0 to 9 which, when multiplied by each other, result in 14.

There aren't that many; an easy way to find the right ones is to divide 14 by each of the numbers from 1 to 9.

Take those two numbers and put the smaller number in the tens place, the larger number in the ones place for the two digit number.

Then take that number and add 45.

The result should be the same two single digit numbers, but in reversed order.

For example, if the question read "Take two numbers whose product is 16 and add 54" you'd multiply through until you got 8x2 and 4x4. Then assume the first number is 28. Add 54, and the result is 82. This is the first two digits in reverse order. It wouldn't work if you'd tried 4x4 by the way, since reversing the digits is the same.

riticool,

**Solution : **Let digit at ten’s place be x

and digit at unit’s place be y

The number = 10x + y

When digits are interchanged, the new number = 10y + x

As per problem,

product of digits = 14 i.e. xy = 14 (1)

Also, 10x + y + 45 = 10y + x

or 10x – x = 10y – y – 45

or 9x = 9y – 45

or x = y-5 (2)

From (1) and (2)

xy = 14

or (y – 5)y = 14

or y2 – 5y – 14 = 0

or y2 – 7y + 2y – 14 = 0

or y(y – 7) + 2(y –7) = 0

or (y + 2) (y – 7) = 0

Either y + 2 = 0 or y – 7 = 0

⇒ y = –2 or y = 7

Rejecting y = –2, we get y = 7

When y = 7, x = 2

The required number = 10 × 2 + 7

= 27

Therefore, the required number is 27

The answer can be solved using a quadratic equation or using matrices, but I gave you a fairly straightforward answer. I hope this is what you needed.

By condtion the product of two digits are 14.

So we assume the digits are x and 14/x. Of course by common sense the choice of the digits have to be(i)2 or 7 or(ii) 7 and 2.

If you add 45 to 27 the result is 72 , which is 27 reversed.

Therefore, **by common sense the solution is 27.**

Aliter:

Let the digits be x in the ten's place and 14/x in units place.

Then the value of the original number: 10x+14/x

If you add 45 to 10x+14/x, we get 10x+14/x+45 which is equal to reversal of digits (14/x x) or 14/x in the ten's place and x in unot's place and whose value is 10(14/x)+x or 140/x +x

Therefore, 140/x +x= 10x+14/x+45 . Multiply by x both sides:

140+x^2 = 10x^2+14 +45x. Or

140-14 = 10x^2-x^2+45x

9x^2+45x- 126 = 0 . Or

x^2+5x-14=0. Or

x^2-2x+7x-14 = 0. Or

x(x-2)+7(x-2) = 0. Or

(x-2)(x+7) = 0.

Therefore, x=2 or x=-7 which is not practical.

So the two digits are 2 and 14/2 or 2 and 7 and the required number is **27.**