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4*9 = 36 so 360 is a possible answer.

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11y ago

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What are the steps to dividing a 2-digit number with a 3-digit number?

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When this 3 digit number is rounded to the nearest ten it rounds to 350 the number is less than 360 and greater than 350 the ones digit is an even number that you count after 2 what is the number?

We will go by the process of elimination to solve this: it is rounded to the nearest ten and gets 350, so it has to be from 345-354. It is greater than 350, so it is 350-354. It is an even number, so it is either 350, 352 or 354. The ones digit is after 2, the answer is 354.


Three digit number that is divisible by 4 and 9?

To be divisible by both 4 and 9 a number must be a multiple of the Least Common Factor of 4 and 9. The LCM is 36. The first 3 digit number in each hundred that is a multiple of 36 are as follows :- 108, 216, 324, 432, 504, 612, 720, 828 and 900.


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Multiply by 8 just like any other number. For example 3 x 8 = 24. To determine if a number is divisible by 8, just consider the last three digits. For example: 25654789008 is divisible by 8 because the last three digits (008) are divisible by 8 (008 / 8 = 1). 78945987345007 is not divisible by 8 because the last three digits are not divisible by 8 (007 / 8 = 0.875).


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To find how many three-digit numbers are multiples of 20, we need to determine the range of three-digit numbers divisible by 20. The smallest three-digit number divisible by 20 is 100, and the largest is 980. To find the count of numbers in this range, we can divide the largest number by 20 and subtract the result of dividing the smallest number by 20, then add 1 to account for the inclusive range. Therefore, the number of three-digit multiples of 20 is (980/20) - (100/20) + 1 = 49 - 5 + 1 = 45.


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The largest two-digit multiple of 5 is 95. This is because the multiples of 5 within the two-digit range start from 10 and go up to 95, with each multiple increasing by 5. The next multiple, 100, is a three-digit number, so 95 remains the largest two-digit multiple.


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What number is it when the the ones digit is one more than its tens digit and it's a composite number has two digits and it's even and has two less than a square number?

Let's list the requirements for the mystery number: One's digit more than the ten's digit. Composite number. Two digits. Even number. Is two less than a square. Now that we've listed the requirements, let's look for a place to "attack" the problem. The easiest is probably to look at squares since our mystery number is two less than a square. Here are the first few squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 and 121. Since our mystery number is 2 less than a square, let's go through that. -1, 2, 7, 23, 34, 47, 62, 79, 98, 119. Now we'll look at two things at once; we'll look only at the 2-digit numbers that are even. (We don't have to look at the "composite number issue" because every even number except 2 is composite since it can be divided by 2.) 34, 62, and 98. Now the number with the one's digit greater than the ten's digit - and there's only one of them: 34. The number 34 has a one's digit greater than the ten's digit, is composite, has two digits, is even, and is 2 less than a square.