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Q: A number is divisible by 4 if the tens and one digits.?

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To know if a number is composite without listing its factors, you can use these rules:All numbers that end with 2, 4, 6, 8, and 0 (even numbers) are divisible by 2.If the sum of the digits in a number is divisible by 3, the number is divisible by 3.If the last two digits are divisible by 4, the number is divisible by 4.All numbers that have 5 in the one's place are divisible by 5.If a number is divisible by 2 AND is divisible by 3, it is divisible by 6.If the last 3 digits of a number are divisible by 8, the number is divisible by 8.If the sum of the digits of number is divisible by 9, the number is divisible by 9.If a number ends wuth 0, the number is divisible by 10.

A number divisible by 123456789 must be 0 or bigger than 123456789. It must, therefore have 1 digit or 9 digits (or more). A remainder of 1 makes no difference to the number of digits. In any case, there can be no number of 4 digits that is divisible by 123456789.

There is no easy test for divisibility by 7, it is often just as quick, if not quicker, to do the division than any devised test. One test I can suggest: 1) split the number into blocks of 3 digits starting from the right hand end (just like when inserting commas to make a number easier to read); 2) Separately sum the units digits, tens digits and hundreds digits of each block by alternately subtracting and adding starting from the right hand end; 3) Add the sum of the units digits to 3 times the sum of the tens digits to twice the sum of the hundreds digits; 4) if the sum in step 4 is divisible by 7, then so is the original number. got 473: 3 +3×7 + 2×4 = 32 which is not divisible by 7, so 473 is not divisible by 7.

No, this number is divisible by 3 and hence is not a prime number. You can tell that fast by adding the individual digits. If that sum is itself divisible by 3, as this one is, then the number itself is divisible by 3.

No. One way to tell is if the sum of a number's digits is divisible by 9, it's divisible by 9. The sum of 7625's digits is 7+6+2+5 = 20

6 is the first number that's divisible by any one of those digits, or all three.

Just think of one. All numbers are divisible. But that's probably not what you wanted to know. It's more likely you want to know how to find what a number is divisible by. For that, it is necessary to familiarize yourself with the rules of divisibility. If the number is even, it's divisible by 2. If the sum of the digits is a multiple of 3, the whole number is divisible by 3. If the last two digits are a multiple of 4, the whole number is divisible by 4. If the last digit is a 0 or a 5, the whole number is divisible by 5. If the number is even and divisible by 3, it's divisible by 6. If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7. If the last three digits are a multiple of 8, the whole number is divisible by 8. If the sum of the digits is a multiple of 9, the whole number is divisible by 9. If the number ends in 0, it's divisible by 10.

65

97 is one example.

A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729.

261, 264, or 267

51, 52, 53, 56, 57, 58, 59

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