970xy

Alright,莫逆家族 let's go step-by-step.

We are told: ( 970xy ) is a 5-digit number(since it's written as digits concatenated: 9, 7, 0, x, y) and it is divisible by 12.

Step 1: Interpret the notation

( 970xy ) means the number is:

[

N = 9 \times 10000 + 7 \times 1000 + 0 \times 100 + x \times 10 + y

]

[

N = 90000 + 7000 + 0 + 10x + y

]

[

N = 97000 + 10x + y

]

where ( x ) and ( y ) are digits, ( 0 \le x \le 9 ), ( 0 \le y \le 9 ).

Step 2: Divisibility by 12

A number divisible by 12 must be divisible by both 3 and 4.

Step 3: Divisibility by 4 rule

A number is divisible by 4 if its last two digits form a number divisible by 4.

Here, the last two digits are ( xy ) (where ( x ) is tens digit, ( y ) units digit).

So ( 10x + y ) must be divisible by 4.

Step 4: Divisibility by 3 rule

A number is divisible by 3 if the sum of its digits is divisible by 3.

Digits: ( 9, 7, 0, x, y )

Sum of digits: ( 9 + 7 + 0 + x + y = 16 + x + y ).

We need ( 16 + x + y ) divisible by 3.

Step 5: Combine conditions

We have:

1. ( 10x + y ) divisible by 4.
2. ( 16 + x + y ) divisible by 3.

Also ( x, y ) digits, ( 0 \le x \le 9, 0 \le y \le 9 ).

Step 6: Try possible ( xy ) divisible by 4

Let’s list two-digit numbers ( 10x + y ) divisible by 4, with ( x ) possibly 0 to 9, ( y ) 0 to 9.

But ( N = 97000 + (10x + y) ), so ( 10x + y ) ranges from 0 to 99.

We can solve by checking ( 16 + x + y \equiv 0 \ (\text{mod} 3) ).

That means ( x + y \equiv 2 \ (\text{mod} 3) ), because ( 16 \equiv 1 \ (\text{mod} 3) ), so ( 1 + (x+y) \equiv 0 \Rightarrow x+y \equiv 2 \ (\text{mod} 3) ).

Step 7: List possible ( xy ) divisible by 4 and ( x+y \equiv 2 \ (\text{mod} 3) )

Let’s list ( 10x+y ) divisible by 4:

Possible endings: 00, 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96.

Now check ( x+y \mod 3 ):

• 00: ( x=0, y=0, x+y=0 \equiv 0 ) → no.
• 04: ( x=0, y=4, x+y=4 \equiv 1 ) → no.
• 08: ( x=0, y=8, x+y=8 \equiv 2 ) → yes.
• 12: ( x=1, y=2, x+y=3 \equiv 0 ) → no.
• 16: ( x=1, y=6, x+y=7 \equiv 1 ) → no.
• 20: ( x=2, y=0, x+y=2 \equiv 2 ) → yes.
• 24: ( x=2, y=4, x+y=6 \equiv 0 ) → no.
• 28: ( x=2, y=8, x+y=10 \equiv 1 ) → no.
• 32: ( x=3, y=2, x+y=5 \equiv 2 ) → yes.
• 36: ( x=3, y=6, x+y=9 \equiv 0 ) → no.
• 40: ( x=4, y=0, x+y=4 \equiv 1 ) → no.
• 44: ( x=4, y=4, x+y=8 \equiv 2 ) → yes.
• 48: ( x=4, y=8, x+y=12 \equiv 0 ) → no.
• 52: ( x=5, y=2, x+y=7 \equiv 1 ) → no.
• 56: ( x=5, y=6, x+y=11 \equiv 2 ) → yes.
• 60: ( x=6, y=0, x+y=6 \equiv 0 ) → no.
• 64: ( x=6, y=4, x+y=10 \equiv 1 ) → no.
• 68: ( x=6, y=8, x+y=14 \equiv 2 ) → yes.
• 72: ( x=7, y=2, x+y=9 \equiv 0 ) → no.
• 76: ( x=7, y=6, x+y=13 \equiv 1 ) → no.
• 80: ( x=8, y=0, x+y=8 \equiv 2 ) → yes.
• 84: ( x=8, y=4, x+y=12 \equiv 0 ) → no.
• 88: ( x=8, y=8, x+y=16 \equiv 1 ) → no.
• 92: ( x=9, y=2, x+y=11 \equiv 2 ) → yes.
• 96: ( x=9, y=6, x+y=15 \equiv 0 ) → no.

Step 8: Valid ( xy ) values

So possible ( xy ) are: 08, 20, 32, 44, 56, 68, 80, 92.

That means ( (x, y) ) = (0,8), (2,0), (3,2), (4,4), (5,6), (6,8), (8,0), (9,2).

Step 9: Check if any other constraints

No, just divisibility by 12. So all these 8 possibilities work.

Step 10: Possibly the problem wants the number of solutions or specific one

If the problem is just “find all possible ( 970xy ) divisible by 12”, we list them:

1. ( 97008 )
2. ( 97020 )
3. ( 97032 )
4. ( 97044 )
5. ( 97056 )
6. ( 97068 )
7. ( 97080 )
8. ( 97092 )

Final answer (if listing numbers):

[

\boxed{97008, 97020, 97032, 97044, 97056, 97068, 97080, 97092}

]

If they asked for how manysuch numbers, it’s ( \mathbf{8} ).