This is the interval problem in primary school mathematics.

According to "5 1 flag inserted on one side of a straight runway with both ends inserted", we can know that there are 50 sections (5 1- 1 = 50) on this runway. Combined with "the interval is 2m", the total runway length can be calculated as 2x50 = 10.

According to "only 26 flags were inserted later", it is known that there are 25 gaps on the runway (26- 1 = 25). Combined with the total runway length 100m, the average gap can be calculated: (100 ÷ 25 = 4m).

Answer process:

2× (51-1) =100 m.

100 ÷ 25 = 4m

A: The average interval is 4 meters.

Extended data:

Law of division:

(1) Divide from the high order of the dividend, and look at the first few digits of the dividend. If it is not enough, look at one more digit.

(2) Write the quotient on the divisor. If the divisor is not enough, put the quotient 0 on it.

(3) The remainder of each division must be less than the divisor. Place the digits of the divisor in the right of the remainder and continue the division.

The essence of division:

The property that the quotient of (1) is invariant is that the dividend and divisor are multiplied or divided by a number (except zero), and the quotient is invariant.

a/b=(a*n)/(b*n)=(a/n)/(b/n)

(2) The sum (difference) of two numbers is divided by a number, which can be used to divide two numbers respectively (in the case of divisibility), and then the sum (difference) of two quotients can be found.

(a+b)/c = a/c+b/c； (a-b)/c=a/c-b/c