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2010

Answer 2

If you half it and you get a multiple of 3, then what you are saying is it is a multiple of 6 (2 x 3 = 6).

So you want a 4 digit number, a number between 1000 and 9999, which is a multiple of 6.

The lowest is 1002 (= 6 x 167), followed by 1008, 1014, 1020 ... etc. up to 9996 (= 6 x 1666).

So, there are a total of 1500 such numbers.

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Q: What is a 4 digit number that is a multiple of 2 and if you take half you get a multiple of 3?

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It doesn't always. It's true only if the 4-digit numbers are mirror images of each other.

Quite simply, there is no rule that states that all multiples of a one-digit number have to end with that same digit, and no reason why it should be so. In the case of 2, and simply because 10 is a multiple of 2, the last digit (of a multiple of 2) is also a multiple of 2 - but it doesn't even have to be 2; it can be 0, 2, 4, 6, 8. Similar with 5, since 10 is also a multiple of 5. In the case of 4, since 100 is a multiple of 4, the last two digits of any multiple of 4 are a multiple of 4. For instance, take the number 4524 - since the last two digits (24) are a multiple of 4, the whole number is. Suggestion: Take some multiples of 4 (or of some other number), and try to look for patterns. Note that for some numbers, the patterns are so complicated, that it's easier to figure out whether a number is a multiple of another by actually doing the division.

Take the smallest 6-digit even number, then subtract one from it.

Impossible. With all multiples of nine the product of its digits is 9. Take, for exampe, 117 or 153.

When rounding to a specific place, take a look at the digit immediately to the right of the target, in this case, the hundreds place. If that digit is 4 or less, zero everything to the right of the target out. If that digit is 5 through 9, increase the target by one and zero everything to the right of it out. If the target is a 9, increasing it will turn it to zero and increase the digit to the left of the target by one.

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The question asserts something that is not true. For example, 2000 - 1000 = 1000. 1000 is not a multiple of 9.

It doesn't always. It's true only if the 4-digit numbers are mirror images of each other.

Any multiple of four must be even; any number ending in three is odd. Look at the multiples of four from 1 to 9. They are in the order : 4,8,12,16,20,24,28,32,36. Now, any number when multiplied by 4, will have the one's digit as the one's digit of the above multiples In the above multiples, no multiple has 3 as the one's digit, i.e., no multiple ends in three. Take an example. 15x4 will have 0 in the one's digit because 5x4 = 20 and has 0 in the one's place.

If you take the units digit and double it then subtract it from the rest of the number then the difference is divisible by seven. You may have to do it several times. 574 57-8=49 So it is a multiple of 7.

Quite simply, there is no rule that states that all multiples of a one-digit number have to end with that same digit, and no reason why it should be so. In the case of 2, and simply because 10 is a multiple of 2, the last digit (of a multiple of 2) is also a multiple of 2 - but it doesn't even have to be 2; it can be 0, 2, 4, 6, 8. Similar with 5, since 10 is also a multiple of 5. In the case of 4, since 100 is a multiple of 4, the last two digits of any multiple of 4 are a multiple of 4. For instance, take the number 4524 - since the last two digits (24) are a multiple of 4, the whole number is. Suggestion: Take some multiples of 4 (or of some other number), and try to look for patterns. Note that for some numbers, the patterns are so complicated, that it's easier to figure out whether a number is a multiple of another by actually doing the division.

Take the smallest 6-digit even number, then subtract one from it.

4- last two digits are a multiple of 4 7-take the last digit and subtract it from the rest of the number and if multiple of 7 (including 0) 8- last three digits are multiple of 8

The number must meet two criteria:Take the original number excluding its last digit and subtract two times the last digit from it. The result must be divisible by 7.Take the original number excluding its last digit and subtract nine times the last digit from it. The result must be divisible by 13.In both cases, you can repeat the "truncate-and-subtract multiple" stage several times, if required.

Impossible. With all multiples of nine the product of its digits is 9. Take, for exampe, 117 or 153.

When rounding to a specific place, take a look at the digit immediately to the right of the target, in this case, the hundreds place. If that digit is 4 or less, zero everything to the right of the target out. If that digit is 5 through 9, increase the target by one and zero everything to the right of it out. If the target is a 9, increasing it will turn it to zero and increase the digit to the left of the target by one.

The answer is vague but correct . Take any prime number and any other number whose combined digit length is 401 and their product will generate a 400 or a 401 digit number.

There are LOTS of them. Here's just one 423452169