How to prove whether any number can be divided by 3? Next I'll show you how to learn whether any number can be divided by 3 Example 1: can 456321 divide by 3? No, add up all its numbers and divide by 3. If the division is complete, then it can divide by 3. If the division is incomplete, then it can't divide by 3. 4 + 5 = 9, 9 + 6 = 15, 15 + 3 = 18, 18 + 2 = 20, 20 + 1 = 21. 21 can divide by 3, so 456321 can divide by 3 Example 2: can 86453 divide by 3? No, add up all its numbers and divide by 3. If it can divide completely, then it can divide by 3. If it can't divide completely, then it can't divide by 3. 8 + 6 = 14, 14 + 4 = 18, 18 + 5 = 23, 23 + 3 = 26.26, so 86453 can't divide by 3 I read the above paragraph on the Internet. I want to know how to prove this principle?

How to prove whether any number can be divided by 3? Next I'll show you how to learn whether any number can be divided by 3 Example 1: can 456321 divide by 3? No, add up all its numbers and divide by 3. If the division is complete, then it can divide by 3. If the division is incomplete, then it can't divide by 3. 4 + 5 = 9, 9 + 6 = 15, 15 + 3 = 18, 18 + 2 = 20, 20 + 1 = 21. 21 can divide by 3, so 456321 can divide by 3 Example 2: can 86453 divide by 3? No, add up all its numbers and divide by 3. If it can divide completely, then it can divide by 3. If it can't divide completely, then it can't divide by 3. 8 + 6 = 14, 14 + 4 = 18, 18 + 5 = 23, 23 + 3 = 26.26, so 86453 can't divide by 3 I read the above paragraph on the Internet. I want to know how to prove this principle?


It is proved that if a number is ABCDE, then the number = a × 10000 + B × 1000 + C × 100 + D × 10 + e × 1 = a × (9999 + 1) + B × (999 + 1) + C × (99 + 1) + D × (9 + 1) + e × 1 = a × 9999 + B × 999 + C × 99 + D × 9 + (a + B + C + D + e) × 1a × 9999 + B × 999 + C × 99



It is proved that the 48th power-1 of 3 can be divided by numbers between 20 and 30


3^48-1=(3^24+1)(3^24-1)=(3^24+1)(3^12+1)(3^12-1)=(3^24+1)(3^12+1)(3^6+1)(3^6-1)=(3^24+1)(3^12+1)(3^6+1)(3^3+1)(3^3-1)
among
3^3-1=26
3^3+1=28



How many palindromes are there in the five digit number where the hundreds are 0? How many palindromes are there in the even number?


If five palindrome number is ab0ba, then a ≠ 0 can be taken as any number from 1 to 9, counting 9 kinds; B can be taken as any number from 0 to 9, counting 10 kinds;
Therefore, the number of palindromes whose hundreds are zero is 9 × 10 = 90;
The number of palindromes is 4 × 10 = 40;
If a is an even number and the number of hundreds is random, then the number of hundreds can be divided into 10 types from 0 to 9, and the number of palindromes is 4 × 10 × 10 = 400