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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

- x + y = 16
- 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.

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

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63? But it has more than two one-digit factors.

The one digit number that has the same number of factors as the number six is three. Six's factors are 1, 2, 3, and 6 while three's factors are 1 and 3. Therefore, both numbers have the same number of factors.

Sure. Any 1-digit number contains only one-digit numbers as factors. Any larger number, on the other hand, contains itself as a factor, so these can be excluded.

it'll definately one '1'

i do not really know because i am the one that asked the quetion.

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63? But it has more than two one-digit factors.

63

63.

The largest one digit number that has three factors is 9. Its factors are 1, 3, and 9.

The one digit number that has the same number of factors as the number six is three. Six's factors are 1, 2, 3, and 6 while three's factors are 1 and 3. Therefore, both numbers have the same number of factors.

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Sure. Any 1-digit number contains only one-digit numbers as factors. Any larger number, on the other hand, contains itself as a factor, so these can be excluded.