According to the meaning of the question, the front of the car uses 8 wheels and each car uses 6 wheels.

There are 40 wheels, so:

40-8=32, that is, there are 32 wheels left besides the front wheel.

32/6=5……2

That is, the remaining 32 wheels can also be assembled into 5 cars, leaving 2 wheels.

So little * * * has 40 wheels, 8 of which are used in the front. He can't assemble a small train with six carriages.

Extended data:

This kind of problem belongs to the remainder problem in mathematics.

The remainder has the following important properties (A, B and C are all natural numbers):?

(1) The absolute value of the difference between the remainder and the divisor is less than the absolute value of the divisor (applicable to the real number field);

(2) Dividend = divisor × quotient+remainder;

Divider = (dividend-remainder) ÷ quotient;

Quotient = (dividend-remainder) divider;

Remainder = dividend-divisor × quotient.

(3) If the remainders of A and B divided by C are the same, then the difference between A and B can be divisible by C. For example, if the remainders of 17 and1divided by 3 are 2, then17-1can be divisible by 3.

(4) The sum of A and B divided by the remainder of C (except that A and B divided by C have no remainder) is respectively equal to the sum of the remainder of A and B divided by C (or the remainder of this sum divided by C).

For example, 23, the remainder of 16 divided by 5 is 3 and 1 respectively, so the remainder of (23+ 16) divided by 5 is equal to 3+ 1=4. Note: When the sum of the remainder is greater than the divisor, the remainder is equal to the sum of the remainder and divided by the remainder of C. ..

(5) The product of A and B divided by C is equal to the product of A and B divided by C (or the product divided by C).

For example, 23, the remainder of 16 divided by 5 is 3 and 1 respectively, so the remainder of (23× 16) divided by 5 is equal to 3× 1=3. Note: when the product of the remainder is greater than the divisor, the remainder is equal to the product of the remainder divided by the remainder of C.

Arithmetic of division

(1) from the high order of dividend;

(2) How many digits is the divisor? Look at the first few digits of the divisor first. If it's not enough, look at one more digit.

(3) In addition to which one, which one to write the business on;

(4) The remainder of each division must be less than the divisor;

(5) If the quotient of 1 is not enough on the dividend after finding the highest bit of the quotient, write 0 on it.