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A three digit number is represented by ABC. It is known that it can be divided by 2, 3 and 5 at the same time, and a + C = 8. What is the three digit number?
eight hundred and ten
A three digit number is represented by ABC. It is known that it can be divided by 2, 3 and 5, and a + C = 8?
Urgent. Want to know how to calculate! Want formula!
He's a multiple of 30
So C = 0
a=8
So this number is 810840870
A three digit number is represented by ABC. It is known that it can be divided by 2, 3 and 5, and a + C = 5?
A + C = 5 and a ≠ 0, and can be divided by 5
Then a = 5, C = 0
Then the number is 510540570
A four digit 8 () 1 () can be divided by 2,3,5 at the same time. What is the possible number of the four digits?
Write down all the possibilities
Let the first number be x and the last one be y
If the sum of 8 + X + 1 + y is a multiple of 3, it can be divided by 3, if y is a multiple of 2, it can be divided by 2, if y is a multiple of 5, it can be divided by 5
So there are 8010831086108910,
A six digit 568abc can be divided by 3, 4 and 5 at the same time. What's the maximum number of this six digit? Please give us a formula
To meet the requirement of 5 first, the requirement of 5 or 0 at the end is the multiple of 4 at the last two digits, so it can be 20 or 40. Finally, to meet the requirement of 3, the maximum is 568940
A six digit 568abc can be divisible by 3, 4 and 5 respectively. What is the minimum of this six digit number
568020
The mantissa that can be divisible by 5 can only be 0, or the mantissa that can be divisible by 4 is at least even, so the mantissa is 0
Then 3 is the key. 568 divided by 3 is the remainder 1, so we can regard it as 1ab0. Because it is the smallest, we can start from a and set a = 0, B = 1 (no way) and B = 2 (yes)
If there is a 1994 digit a which can be divided by nine, the sum of its digits is a, the sum of its digits is B, and the sum of its digits is C, then C = ()
I need it urgently!
9 yes, I did
If the three digits of a three digit number are a, B, C, and (a + B + C) can be divisible by 9
It is proved that if the hundred digit is a, the ten digit is B, and the single digit is C, then the three digit is 100A + 10B + C = 99A + 9b + (a + B + C), ∵ 99A, 9B, (a + B + C) can be divisible by 9, and the three digit must be divisible by 9
If the three numbers of a three digit number are a, B and C respectively, and a + B + C can be divisible by 9, it is proved that the three digit number must
Prove that this three digit number must be divisible by 9.
The three digits are:
100a+10b+c
=99a+9b+(a+b+c)
=9(11a+b)+(a+b+c)
The above two parts can be divided by 9, indicating that the above formula can be divided by 9
A 1994 digit number a can be divided by 9. If the sum of its digits is a, the sum of its digits is B, and the sum of its digits is C, what is C?
Let the number of each bit of a multi digit number be 9, then a = 1994 × 9 = 17946, B = 1 + 7 + 9 + 4 + 6 = 27, C = 2 + 7 = 9
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