What two digit number has two one digit factors that have the sum of 16 and the difference of 2?

Answer 1

63.

Let the smaller factor be n-1.

Then with a difference of 2, the larger factor is (n-1) + 2 = n+1

The sum of these is:

(n-1) + (n+1) = 2n = 16 ⇒ n=8

Thus the factors are

• 8-1 = 7
• 8+1 = 9

and the number is 7x9 = 63.

(The complete list of factor pars for 63 is: 1 & 63, 3 & 21, 7 & 9.)

Alternatively, if you don't want the simultaneous equations "hidden":

Let the two factors be x & y (so that the number is xy) with x bigger than y. Then

1. x + y = 16
2. x - y = 2

Adding the two equations gives:

2x = 18

⇒ x = 9

and substituting in equation 1 gives:

9 + y = 16

⇒ y = 7

and substituting in equation 2 to check:

9 - 7 = 2 as required.

So the factors are 7 & 9 and so the number is xy = 7x9 = 63.