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If it's to be divisible by 5, it must end in 5 or 0. And if it's to be divisible by 3, adding the digits together must result in a multiple of 3.

While some numbers that end in zero will still work, you can guarantee that the number won't be divisible by four by choosing a number that ends in five.

So you'd need a four-digit number, that ends in 5, whose digits add up to a multiple of 3. The smallest such number would be 1005, and the largest would be 9975.

(However, the actual largest number that would fit your question is 9990; remember that I did say some numbers that end in zero would still work.)

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13y ago
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12y ago

505, 515, 525, 535, 545, 555, 565, 575, 585, 595 are all such numbers.

(Unless you consider 000, 010, 020, 030, 040, 050, 060, 070, 080, 090 three digit numbers)

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Q: What is a four digit number divisible by 3 and 5 but not 4?
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Can you divide 2 3 5 6 9 10 into 305?

305 can only be [exactly] divided by 5 and not 2, 3, 6, 9 nor 10.Divisibility tests:2: NoNumber must be even (last digit divisible by 2). Last digit is 5 which is not even, so the whole number (305) not divisible by 2.3: NoAdd up the digits of the number; if this sum is divisible by 3, then the original number is divisible by 3: 3 + 0 + 5 = 8 which is not divisible by 3, so the original number (305) is not divisible by 3.5: YesLast digit is 0 or 5. Last digit is 5, so 305 is divisible by 5. 6: No6 = 2 x 3, so number must pass both tests for divisibility of 2 and 3; 305 fails tests for (both) 2 and 3 (above), so is not divisible by 6. 9: NoAdd up the digits of the number; if this sum is divisible by 9, then the original number is divisible by 9: 3 + 0 + 5 = 8 which is not divisible by 9, so the original number (305) is not divisible by 9.10: NoThe last digit of the number must be 0. Last digit is 5, so 305 is not divisible by 10.


What three digit number is divisible by 3 and 5?

36360 is a 5 digit number that is divisible by both 3 and 9.


How can you tell if a number is divisible by 12?

If a number is divisible by both three and four, it's divisible by twelve.


How do you use divisibility rules to find at least for factors of a number?

Take the number 3336. You know it's divisible by 1 because everything is. You know it's divisible by 2 because it's even. You know it's divisible by 3 because the digits add up to a multiple of 3 and you know it's divisible by 4 because the last two digits are divisible by 4. So you've found at least four factors: 1,2,3 and 4.


Is 30030003 divisible by 8 or 9?

It is divisible by 8 because 202008/8 = 25251 But divided by 9 it will leave a remainder ----------------------- To test if a number is divisible by 8 add the ones digit to twice the tens digit to 4 times the hundreds digit; if this sum is divisible by 8 then so is the original number. By repeating the test on the sum until a single digit remains, only if this single digit is 8 is the original number divisible by 8, otherwise it gives the remainder when divided by 8 (except if it is 9, in which case the remainder is 1 - the excess of 9 over 8). For 202008: ones_digit + 2 × tens_digit + 4 × hundreds_digit = 8 + 2 × 0 + 4 × 0 = 8; so 208008 is divisible by 8. To test if a number is divisible by 9, add up the digits of the number; if this sum is divisible by 9, then so is the original number. By repeating the test of the sum until a single digit remains, only if this single digit is 9 is the original number divisible by 9, otherwise it gives the remainder when the original number is divided by 9. (This single digit is also called the "digital root" of the number.) For 202008: 2 + 0 + 2 + 0 + 0 + 8 = 12; 1 + 2 = 3; so 202008 is not divisible by 9; it has a remainder of 3 when divided by 9.