Add three numbers after 865 to form a six digit number, so that it can be divided by 3, 4 and 5 respectively, and make the value as small as possible, then the six digit number is______ ．

Add three numbers after 865 to form a six digit number, so that it can be divided by 3, 4 and 5 respectively, and make the value as small as possible, then the six digit number is______ ．

Can be divided by 4, 5, the number of bits must be 0, can be divided by 4, the last two bits can be divided by 4, so the number of ten must be even, can be divided by 3, the sum of each number can be divided by 3, 8 + 6 + 5 = 19, at least 2 can be divided by 3, so the minimum number is 865020

A six digit () 1993 () is divisible by 44
Find a six digit () that can be divided by 44

44=4×11
If you can divide by 4, you have to divide by 4 and 11 at the same time
Divisible by 4: the last two digits can be divisible by 4
Features divisible by 11: odd digit sum, difference from even digit sum, divisible by 11 or 0
If the last two digits can be divided by 4, then the last one is 2 or 6
9+3=12
1+9=10
12-10=2
The first one is two times smaller than the last one
So the last one is 6 and the first one is 4
The six digits are: (4) 1993 (6)

Six digit 283a6b. When a is () and B is (), the six digit can be divided by 44

Consider 44 = 11 * 4, then 6B should be divisible by 4, which can be 60, 64, 68, in addition, 2 + 3 + 6-8-a-b should be divisible by 11,
That is to say, a + B-3 can be equal to 0, 11. That is, a + B = 3 or 14. Combined with B, which may be 0, 4 or 8, we can get three answers: 3, 0, 7, 4, 3 and 8, namely 283360, 283764 and 283368

If a six digit * 1993 * can be divided by 33, what is the six digit?

The number that can be divisible by 33 is divisible by both 3 and 11. If it is divisible by 3, the sum of six digits is a multiple of 3, that is, if the unknown two digits are added and divided by 3, if the remaining two are divisible by 11, the sum of odd digits and even digits is equal

What is the maximum number of six digits that can be divisible by 225 if you fill in the number in

225=3^2×5^2 = 9×25
If the number is divided by 25, the last two digits are divided by 25, and the last two digits can only be 25
If 20 □ 025 is divisible by 9, then the sum of each digit is divisible by 9,
□ + 2 + 2 + 5 = □ + 9 divisible by 9
Maximum 9
The maximum number is 209025

Fill in the appropriate number in (), so that six digits () 1998 () can be divided by 56

Can be divided by 56, that is, can be divided by 7, 8
If you want to be divided by 8, the last three places will be divided by 8
98x is divisible by 8, obviously x = 4
X19984 to be divisible by 7, according to the last cut three method
984 - X19
= 984-19 - 100X
= 965 - 100X
= 138*7 - 14*7X - 1 - 2X
= 7*(138+14X) - (2X+1)
That is to say, 2x + 1 can be divided by 7, and X obviously = 3
In conclusion, 319984 can be divided by 56

What is the number of six digit 20 * * 08 divisible by 49 * *

The original number plus a multiple of 49:
98*4=392
have to
20*(*+4)00
Subtract a multiple of 49 from the front:
98*2=196
have to
(*+4)(*+4)00
If this number is divisible by 49, the first two numbers are 49 or 98
So the original number is
200508 or 205408

A2875b can be divided by 99 to find out what a and B are

Divisible by 99, divisible by 9 and 11
If it is divisible by 9, then a + 2 + 8 + 7 + 5 + B = 22 + A + B can be divisible by 9
So a + B = 5 or 14
If it can be divided by 11, the sum of even digits minus the sum of odd digits can be divided by 11
(2 + 7 + b) - (a + 8 + 5) = b-a-4 can be divided by 11
A = - 4 or B = - 7
a+b=5,b-a=-7,b=-2

For a five digit number, the first three digits are a complete square number, the last two digits are the same, and the five digit number can be divided by 99. How many five digits are there?

If the first three digits are a perfect square, let it be x ^ 2, and the last two are the same, let it be 10A + a = 11a
This number is expressed as a: a = 100x ^ 2 + 11a
This number can be divided by 99 and 11. 11a can be divided by 11. So 100x ^ 2 can also be divided by 11, that is, x ^ 2 can also be divided by 11. Let x ^ 2 = (11b) ^ 2 = 121b ^ 2
A=100*121b^2+11a
121b ^ 2 is a three digit number, so B can only be 1 and 2
If B = 1, a = 100 * 121 + 11a = 12100 + 11a
Because a can be divisible by 99, it can also be divisible by 9. The characteristic of being divisible by 9 is that the sum of all numbers can also be divisible by 9. Then the sum of all numbers of a is: 1 + 2 + 1 + A + a = 4 + 2A
4 + 2A can't be 9, so it can only be 18. A = 7, so this number can be 12177
If B = 2, a = 121 * 400 + 11a = 48400 + 11a
The sum of your numbers is: 4 + 8 + 4 + A + a = 16 + 2A
The maximum is 34, but it can't be 27, because it's an even number, so it can only be 18, at this time a = 1
A=48411
Well, it could be 12177 and 48411

A five digit. 3ab98 can be divided by 11 and 9, then the five digit is______ ．

∵ five digits. 3ab98 can be divided by 11 and 9, ∵ 3 + A + B + 9 + 8 = 20 + A + B = 9x, 3 + B + 8-a-9 = B-A + 2 = 11y, and ∵ 0 ≤ a ≤ 9, 0 ≤ B ≤ 9, ∵ 0 ≤ a + B ≤ 18, 0 ≤ B-A ≤ 9,