How many 4-digit numbers are divisible by 3 and have 23 as their last two digits?? Please show work/explain!!!

Question

How many 4-digit numbers are divisible by 3 and have 23 as their last two digits?? Please show work/explain!!!

Answer 1

This method is a bit long, but it should still work...

Any number that is divisible by 3 has the property that the sum of its digits is also divisble by 3.
Ex. 15 is divisible by 3 and 1 + 5 = 6 is divisible by 3.

Therefore, for a 4 digit number XY23 to be divisible by 3, (X + Y + 2 + 3) must be divisible by 3.

so X + Y + 2 + 3 could = 6 or 9 or 12 or 15 or 18 or 21 (24 gets too big.. see next line)

which means X + Y could = 1 or 4 or 7 or 10 or 13 or 16 (21 is too big for the sum of 2 single digit numbers)

for a sum of 1 you could have (1,0) = 1 choice

for a sum of 4 you could have (1,3) or (2,2) or (3,1) or (4,0) = 4 choices

for a sum of 7 you could have (1,6) or (2,5) or (3,4) or (4,3) or (5,2) or (6,1) or (7,0) = 7 choices

keep going with 10, 13, 16 and all up all the choices...

Answer 2

Any number that ends in 5 or 0 is divisible by 5. Any number with the last two digits divisible by 4. The First number that is divisible by both 4 and 5 is 120.

Answer 3

Since 3 is divisible by 3, this translates into how many three digit numbers are divisible by 3 and end in 2.

Again, that means that the first two digits must divide by 3 giving a remainder of 1.

There are 30 two digit numbers like this.

The full list is:

1023
1323
1623
1923
2223
2523
2823
3123
3423
3723
4023
4323
4623
4923
5223
5523
5823
6123
6423
6723
7023
7323
7623
7923
8223
8523
8823
9123
9423
9723