Let n be a natural number between 100 and 200, then there are several n n that satisfy 7n 2 which is

c 207n+2=5k (k=1,2,3,..

then k=(7n+2) 5=n+(2n+2) 5 ream(2n+2) 5=p (p=1,2,3,..

then n=(5p-2) 2=2p+(p-2) 2 ream(p-2) 2=m then p=2m+2 (m=1,2,3...

n=2p+(p-2)/2=2(2m+2)+(2m+2-2)/2=4m+4+m=5m+4

100100<5m+4<200

The corresponding n and 7n+2 are respectively.

n1=5*20+4=104 7n+2=7*104+2=730n2=5*21+4=109 7n+2=7*109+2=765n3=5*22+4=114 7n+2=7*114+2=800n20=5*39+4=199 7n+2=7*199+2=1395

Two out of every ten numbers are divisible by 5, i.e. 4 and 9

2 out of 100-110.

110 out of 120-2.

There are 2 in 120-130, so there are 2*10=20 in total.

n must not be a 0 for the core.

Since 2 (n+4)-2 n = 16 * 2 n -2 n = 15 * (2 n).

When n is not 0, 2 n must be a multiple of 2, so 15*(2 n) must be a multiple of 30 times of Hui's search calendar. Certification.

Summary. Natural numbers 1-700, there are 460 numbers that are multiples of and 7 1 700, multiples of 2 have 700 2 = 350 (pieces), multiples of 5 have 700 5 = 140 (pieces), multiples of 7 have 700 7 = 100 (pieces), multiples of 2 and 5 have 700 10 = 70 (multiples of 2 and 7 have 700 14 = 50 (pieces), multiples of 5 and 7 have 700 35 = 20 (pieces) 2 and multiples of 5 and 7 have 700 70 = 10 (pieces), so 2 or 5 or 7 The multiples of 350 + 140 + 100 - (70 + 50 + 20) + 10 = 460 (pieces) Answer: There are 460 numbers that are multiples of and 7.

Natural numbers 1-700, there are 460 numbers that are Multiples of 7 1 700, multiples of 2 have 700 2=350 (pcs), multiples of 5 have 700 5=140 (pcs), multiples of 7 have bumps to do 700 7=100 (pcs), multiples of 2 and 5 have 700 10 = 70 (multiples of 2 and 7 have 700 14 = 50 (pcs), multiples of 5 and 7 have 700 35 = 20 (pcs) Multiples of 2 and 5 and 7 have 700 70 = 10 (pcs), So there are multiples of 2 or 5 or 7 350 + 140 + 100 - (70 + 50 + 20) + 10 = 460 (pieces) Answer: There are 460 numbers that are multiples of and 7.

Subtract the double-counted.

Any number n can be divided by the remainder of 6, which can be divided into six categories: 0, 1, 2, .,Yu 5Namely.

n＝6n;n＝6n+1;

n＝6n+2;

n＝6n+5;

Of the six classes above, n 6n; n＝6n+2;n＝6n+3;n 6n+4 is a multiple of 2, or a multiple of 3. Thus, by the inscription, n panicle trembling sedan 6n + 1; n＝6n+5;In line with the topic, it is not a multiple of 2, nor is it a multiple of 3.

1) At n 6n+1.

n 2-1 = (n 1) (n 1) = 6n + 2) * 6n so, 6 n 2 + 5

2) At n 6n + 5.

n^2-1=（n＋1）（n－1） =6n+6)*（6n＋4）＝6（n+1）*（6n＋4）

So, 6 n 2+5 n 2-1 6 6(n + 1)*(6n 4)+6

First of all, we need to make sure that there must be a multiple of 2 for every 2 consecutive numbers, and the same is true for 5 and 7. So:

Multiples of 2: 700 2=350 (pcs).

Multiples of 5: 700 5 = 140 (pcs).

Multiples of 7: 700 7 = 100 (pcs).

So: in the natural number 1 700, there are (350 + 140 + 100) numbers that are multiples of or 7, which is 590.

Of the 1,700 natural numbers, 241 are multiples of or 7. Among them, there are 233 multiples of 2, 140 multiples of 5, and 70 multiples of 7.

The multiples are: 1000 5 = 200 (pieces), the multiples of 7 are: 1000 7 Gao Panxun 142 (pieces), and the multiples of 35 are double-calculated:

1000 (5 7) 28 (pcs), so it is 200 + 142-28 = 314 (pcs).

Answer: There are 314 numbers that are multiples of 5 or have an approximation of 7;

So the answer is: 314

200-([200 5 ]+200 8 ]-200 5 8 ])200-[40]-[25]+[5],=200-40-25+5,=140 (pcs) Answer: There are 140 multiples of potato pies that are neither 5 nor 8 so the answer is: 140

3n+22n+3

3n2n+1

3n+23n

2n+32n+13n

1) Collapse shouts -2n

10×（3n2n

So 3n+2

2n+33n

2n+1 must be a multiple of Qingpeng 10