Q: What is a 4 digit number that is divisible by 8?

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If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.

Yes. To check if a number is divisible by 4, if the last 2 digits (as a number) are divisible by 4, so is the original number. 32 = 4 x 8 so as 32 is divisible by 4, so is 211032. An alternative test for 4: add the last digit to twice the preceding digit, and if this sum is divisible by 4 so is the original number. This can be repeated until a single digit remains which must be 4 or 8 (to be divisible by 4) if the original number is divisible by 4. for 211032: 2 + 3x2 = 8 → [single digit 8 which] is divisible by 4, so 211032 is also divisible by 4.

4000

2400

1000.

No. To test if a number is divisible by 8: * first all multiples of 8 are even, so the number must be even; * then: add 4 times the hundreds digit to twice the tens digit to the ones digit - if this sum is divisible by 8, then so is the original number. As the test can be applied to the sum, repeating this summing until a single digit remains, only if this single digit is 8 is the original number divisible by 8. For 100: 4x1 + 2x0 + 0 = 4 which is not 8, so 100 is not divisible by 8.

2400

5040

4444

Yes. To check divisibility by 8 add the units digit to twice the ten's digit to four times the hundred's digit; if this sum is divisible by 8, so is the original number: 4 + 2x4 + 4x3 = 4 + 8 + 12 = 24 which is divisible by 8, so 110344 is also divisible by 8.

10001000/8 = 125

There is no 5 digit number which is divisible by 23456910 (an 8-digit number)..

No. All multiples of 8 are even, but 141 is odd, so 141 is not divisible by 8. ------------------------------------------ To test if a number is divisible by 8, add 4 times the hundreds digit to 2 times the tens digit to the ones digit; if this sum is divisible by 8, then so is the original number. The test can be repeated on the sum, so continue the summing process until a single digit remains - only if this single digit is an 8 is the original number divisible by 8. 141 → 4 × 1 + 2 × 4 + 1 × 1 = 13 13→ 4 × 0 + 2 × 1 + 3 = 5 5 is not 8, so 141 is not divisible by 8 141 ÷ 8 has a remainder of 5.

It would be 9,992 since 1,249 * 8 = 9992. 1250 * 8 would equal 10,000 and this is not a 4 digit number, but a 5 digit number. The divisibility rule for 8 states that the last 3 digits of a number must be divisible by 8 and this is true as 992/8 = 124. Hopefully that helps.

If the number formed by the last three digits is divisible by 8. This requires that: if the digit in the hundreds place is even, the last two digits must form a number divisible by 8 and if the digit in the hundreds place is odd, the last two digits must form a number divisible by 4 but not by 8.

i do not know but i think it 480

7200 is one possible answer.

2 x 6 + 0 = 12 2 x 1 + 2 = 4 4 is not [divisible by] 8, so 60 is not divisible by 8. (The remainder when 60 is divided by 8 is 4). To test divisibility by 8: Add together the hundreds digit multiplied by 4, the tens digit multiplied by 2 and the units (ones) digit. If this sum is divisible by 8 so is the original number. (Otherwise the remainder of this sum divided by 8 is the remainder when the original number is divided by 8.) If you repeat this sum on the sum until a single digit remains, then if that digit is 8, the original number is divisible by 8 otherwise it gives the remainder when the original number is divided by 8 (except if the single digit is 9, in which case the remainder is 9 - 8 = 1).

Yes. If the last digit is even (ie one of 0, 2, 4, 6, 8) then the whole number is divisible by 2. The last digit is 4, so the number is divisible by 2.

If it is a multiple of 4. To check if a number is divisible by 4, first it must be even (end in one of the digits 0, 2, 4, 6, 8). Then add twice the tens digit to the ones digit; if this sum is divisible by 4 then so is the original number. As the test can be applied to the sum, repeat the summing until a single digit remains; if this digit is 4 or 8 then the original number is divisible by 4. eg 123456789 is not divisible by 4 as it is odd (ends in one of the digits 1, 3, 5, 7, 9). eg 123456798: 2x9 + 8 = 26; 2x2 + 6 = 10; 2x1 + 0 = 2 which is not 4 nor 8, so 123456798 is not divisible by 4. eg 123987564: 2x6 + 4 = 16; 2x1 + 6 = 8 which is 4 or 8, so 123876564 is divisible by 4.

Add 4 times the hundreds digit to twice the tens digit to the units digit. If this sum is divisible by 8, then the original number is divisible by 8. If this summing is repeated until a single digit remains, if this digit is 8, then the original number is divisible by 8 otherwise it gives the remainder of the original number divided by 8 (unless it is 9, in which case the remainder is 9 - 8 = 1). 7 x 4 + 9 x 2 + 6 = 28 + 18 + 6 = 52 5 x 2 + 2 = 12 1 x 2 + 2 = 4 4 is not 8, so 796 is not divisible by 8 (the remainder is 4).

No. If the sum of (the hundreds digit times 4) plus (the tens digit times 2) plus (the ones digit [times 1]) is divisible by 8, then so is the original number: 2 x 4 + 1 x 2 + 8 x 1 = 18 which is not divisible by 8, so 218 is not divisible by 8.

To check for divisibility by 8 add four times the hundreds digit to twice the tens digit to the units digit; if this sum is divisible by 8 then so is the original number: For 728: 4×7 + 2×2 + 8 = 40 which is divisible by 8, so 728 is divisible by 8. The test can be repeated on the sum, so if the test is repeatedly applied to the sums until a single digit remains, only if this single digit is 8 then the original number is divisible by 8 (otherwise the excess of this number over a multiple of 8 is the remainder when the original number is divided by 8, ie if it is less than 8 it is the remainder, else it is 9 and the remainder is 9 - 8 = 1). For 728: 4×7 + 2×2 + 8= 40 40: 4×0 + 2×4 + 0 = 8 This single digit is 8, so 728 is divisible by 8.

No; as 4 is less than 7321694 it cannot be divisible by it. Nor is 7321694 divisible by 4. There are two tests for divisibility by 4 that I can offer. First (test given in most books): If the last two digits of the number (as a number) are divisible by 4 then so is the original number. Second (much easier test): add twice the tens digit to the ones digit; if this sum is divisible by 4 then so is the original number. Using second test, for 7321694: 2x9 + 4 = 22 which is not divisible by 4, so 7321694 is not divisible by 4. The second test can be repeated until a single digit remains; if this single digit is 4 or 8, then the original number is divisible by 4. otherwise not: 7321694 → 2x9 + 4 = 22 22 → 2x2 + 2 = 6 → not 4 or 8, so original number is not divisible by 4.

There is no answer because for a number to be divisible by 8, it would end in an even number and therefore be divisible by 2.

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