A two-digit number is such that the product of its digits is 35. If 18

1.	A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.
Answer» Let the two digit number is xy = 10x + y. Given that product of required number is 35. i.e., xy = 35 … (1) (∵Unit digit of number is y and second digit of number is x.) Also, Given that if 18 added to number (10x + y), The digits interchange their places i.e., It becomes yx = 10y + x. i.e., 10x + y + 18 = 10y + x ⇒ 9x − 9y + 18 = 0 ⇒ 9(x − y) = −18 ⇒ y − x = 2 … (2) Now, (y + x)² = (y − x)² + 4xy = 4 + 4×35 (using equation (1) & (2)) = 4 +140 = 144 ⇒ y + x = 12 … (3) Now, Adding equation (2) and (3), we get (y – x) + (y + x) = 2 + 12 ⇒ 2y = 14 ⇒ y = 7. Now, Putting y = 7 in equation (3), we get 7 + x = 12 ⇒ x = 12 – 7 = 5. ∴ The required number is xy = 10x + y = 10 × 5 + 7 = 57.

A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Answer»

Let the two digit number is xy = 10x + y.

Given that product of required number is 35.

i.e.,

xy = 35 … (1)

(∵Unit digit of number is y and second digit of number is x.)

Also,

Given that if 18 added to number (10x + y),

The digits interchange their places i.e.,

It becomes yx = 10y + x.

i.e., 10x + y + 18 = 10y + x

⇒ 9x − 9y + 18 = 0

⇒ 9(x − y) = −18

⇒ y − x = 2 … (2)

Now,

(y + x)² = (y − x)² + 4xy = 4 + 4×35

(using equation (1) & (2))

= 4 +140 = 144

⇒ y + x = 12 … (3)

Now,

Adding equation (2) and (3), we get

(y – x) + (y + x) = 2 + 12

⇒ 2y = 14

⇒ y = 7.

Now,

Putting y = 7 in equation (3), we get

7 + x = 12

⇒ x = 12 – 7 = 5.

∴ The required number is xy = 10x + y

= 10 × 5 + 7

= 57.

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