Here is a intellectual math problem?

Question

T and U are two digits of the number 9T68U. This 5 digit number is divisible by 15. Find all of the different possible values of (T+U).

Answer 1

9 + t + 6 + 8 + u = 23 + t + u

since the number is divisible by 15, it has to be divisible by 3 and 5.

hence, u is either 0 or 5
if u is 0, then 23 + t + u = 23 + t

23 + t should be divisible by 3 (if the original number is divisible by 3)

Since 23 leaves a remainder 2 when divided by 3, this means t is a number between 0 and 9 that leaves a remainder of 1.

Hence, if u = 0, then t has to be 1 or 4 or 7

If u = 5, then 23 + t + u = 28 + t
and since 28 leaves a remainder 1, t would have to leave remainder 2 when divided by 3.
Thus, t = 2 or 5 or 8

Hence t + u = 1 or 4 or 7 or 10 or 13.

Answer 2

Simpler answer.

A number is divisible by 15 if and only if it is divisible by 3 and 5.

A number is divisible by 3 if and only if the sum of its digits are divisible by 3:

9 + T + 6 + 8 + U = 23 + (T+U) must be divisible by 3.

A number is divisible by 5 if and only if its last digit is 0 or 5:

U must be 0 or 5.

T+U can be between 0 and 14 since U is at most 5 and T is at most 9. Every value of T+U can be achieved in this range with U=0 or 5.

The values of T+U between 0 and 14 which, such that T+U+23 is divisible by 3 are those where T+U-1 is divisible by 3. (Since T+U-1+24 is divisible by 3 exactly when T+U-1 is.) These values are 1, 4, 7, 10, 13.

This yield possible values

T+U = 1 (T=1; U=0)
T+U = 4 (T=4; U=0)
T+U = 7 (T=7; U=0)
T+U = 10 (T=5; U=5)
T+U = 13 (T=8; U=5)

Note, the problem didn't ask to find all solutions, but all the possible values of T+U. These values are the only possible values for T+U since U=0 or 5 means that T+U<14, and T+U-1 divisible by 3 means that T+U must be one of these numbers.

Answer 3

The prime factors of 15 are 3 and 5. So the number 9T68U must be divisible by 3 and 5.

The test for divisibility by 5 is that the last digit is 0 or 5.

The test for divisibility by 3 is that all the digits add to a multiple of 3. So far 9 + 6 + 8 = 23. So you need to add 1, 4, 7, 10, etc.

If U is 0, then T will be 1, 4 or 7
If U is 5, then T will be 2, 5 or 8

The question asked for the possible values of (T + U). This would be:
1, 4, 7, 10 or 13

Answer 4

The number is divisible by 15, so it is also divisible by the factors of 15, i.e., 3 and 5. Now think of the divisibility tests for 3 and 5.

Answer 5

t u
2 5
5 5
8 5
1 0
4 0
7 0

Answer 6

15 IS DIVISIBLE BY 5 AND 3.AS U IS THE LAST DIGIT OF THIS NUMBER,IT WILL BE DIVISIBLE BY 5.NOW,ADD ALL SUCH NUMBERS TO THOSE VALUES SUCH THAT (T+U)+9+6+8 IS DIVISIBLE BY 3.

Answer 7

T ... U
1 ... 0
2 ... 5
4 ... 0
5 ... 5
7 ... 0
8 ... 5