r \u 5× 1/3 = r \u 5 \u 3 = r \u 15

Then: the minimum r is15;

Similarly: 6/7× s ÷ 0.6 = s× 30 ÷ 21= s×10 ÷ 7.

So: s is at least 7.

2.

M \(M/4+M/6)= M \( 10M/24)= 24/ 10 = 2.4

(3/A- 1/2A)×A = 3- 1/2 = 2.5

3.

2 1/4=42/8=63/ 12= …

That is, the following numbers are 4,8,12 ...; Their sum is 2 1, 42, 63 …

If the original number is 5, then:

1+2+3+4+5= 15,

If it fails to reach 2 1, give up;

If the original number is 9, then:

1+2+3+4+5+6+7+8+9=45,

Take away 3, and the average value of the remaining four numbers is 42/8 = 21/4;

If the original number is 13, then:

1+2+3+4+5+6+7+8+9+ 10+ 1 1+ 12+ 13=9 1,

We found that 9 1-63=28, and there was no satisfier in the number taken, so we discarded this situation.

Similarly, when the primitive numbers are 17 and 2 1…,

Because the denominator value is getting bigger and bigger, taking only one number is not enough to make the average of other numbers reach 2 1/4.

So I won't consider this situation afterwards.

So this string of natural numbers is: 1-9, and the erased number is 3.