How many 4-digit numbers are divisible by 3 and have 23 as their last two digits??

Answer

Let X be your 4 digit number.
X = ‘ab23’ = 1000a + 100b + 2*10 + 3
= ( a + b + 2 ) + 999a + 99b + 21
= ( a + b + 2 ) + 3 x ( 333a + 33b + 7 )

Thus, if 3 divides X we must have 3 divide ( a + b + 2 ). In other words, ( a + b + 2 ) is a multiple of 3, say 3k where k is a positive integer.

Thus, ( a + b + 2 ) = 3k. But note that since a & b are both digits of a numeber in base 10, thus they belong to the interval [0,9]. That is
( a + b + 2 ) > 0 + 0 + 2 = 2, that is ( a + b + 2 ) > 2. and
( a + b + 2 ) < 9 + 9 + 2 = 20, that is ( a + b + 2 ) < 20

Since ( a + b + 2 ) = 3k , we get.,
2 < 3k < 20, or , 2/3 < k < 20/3 < 7.
Hence the only “integer” possibilities for k are
k = 1,2,3,4,5,6.
We get ( a + b + 2 ) = 3, 6, 9, 12, 15, 18 or equivalently,
( a + b ) = 1, 4, 7, 10, 13, 16.
So for “each” these “equations”, how many solutions are there? The number of solutions for the (a+b=above) is the number of 4 digit numbers that are divisible by 3 and have 23 as their last two digits… How do we proceed?

We will use a “theorem”.

For x(1) + x(2) + … x(k) = m. where k & m are positive integers, the number of solutions in positive integers is :
C( m – 1, m – k ) = (m -1)! / [( m – k )! * ( k – 1)! ]. Note that x(1) is variable 1, x(2) is variable 2 etc.

And, n! = n x (n -1) x (n -2)x … 2 x 1

The “theorem” above only deals with “positive integers” in its indicies but we have “zero as a possible answer as well”.

Therefore we find first “positive integer solutions” and count the number “solutions which have one of a,b, = 0” later.

For a + b = 1. Since a cannot be zero ( first digit ) thus only solution is (a,b) = (1,0). One answer here.

Number of positive integer solutions for a+b= 4 means that in the above, k = 2 & m = 4. Thus the number of solutions is just = C( m – 1, m – k ) = C( 3, 2 )= (m -1)! / [( m – k )! * ( k – 1)!
=[ 3! ] / [ 2! * 1! ] = 3. That is there are 3 solutions in positive integers implying 3 three digit numbers in this case.

We proceed “similarly” for ( a + b ) = 7, 10, 13, 16 and get,
The number of 3 digit numbers
= C ( 6 , 5 ) + C ( 9 , 8 ) + C ( 12 , 11 ) + C ( 15 , 14 )
= 6 + 9 + 12 + 15
= 42.

Hence, we have for positive integers a & b, and ( a + b ) = 1, 4, 7, 10, 13, 16.
We have a total of
= 1 + 3 + 42 = 46 integers that satisfy our problem.

What about the case where ” one of a & b = 0 “? Since the “theorem” above is only valid for the number of “positive integer solutions”. Indeed consider
( a + b ) = 1, 4, 7, 10, 13, 16 again.
If one of a or b is = 0, since a CANNOT be zero ( first digit of an integer ), this b = 0. But if b = 0, the equations above yields,
a = 1, 4, 7, 10, 13, 16. Since ‘a’ is a digit of a number it can only assume values between 1 to 9. Thus a = 1,4,7 for b = 0 are the only “valid” answers here. Since we have “counted” for the answer pair (a,b) = (1,0) earlier [ see above ] , Thus in this case we have “two new” solutions that is (a,b) = (4,0) and (7,0). That is two answers for this case.

Summing up, we can conclude that
the “total” number of integers that satisfy the condition of the problem = 46 + 2 = 48 integers.