Oh, isn't that a happy little question! To find a number divisible by 525, we can start by looking for numbers that are multiples of both 5 and 105 (which is 525 divided by 5). So, any number that is a multiple of 525, like 525, 1050, 1575, and so on, will be divisible by 525. Keep on exploring and you'll find the perfect number that fits just right!
525 is divisible by 5, since it ends with a 5. 525/5 = 105 525 is not divisible by 2, since it does not end with 0, 2, 4, 6, or 8. 525/2 = 262.5
Yes, evenly
510
The list is infinite. Here are some of them: 210, 315, 420, 525, 630, 735, 840.
1, 3, 5, 7, 15, 21, 25, 35, 75, 105, 175, 525
Oh, isn't that a happy little question! To find a number divisible by 525, we can start by looking for numbers that are multiples of both 5 and 105 (which is 525 divided by 5). So, any number that is a multiple of 525, like 525, 1050, 1575, and so on, will be divisible by 525. Keep on exploring and you'll find the perfect number that fits just right!
525 is divisible by 5, since it ends with a 5. 525/5 = 105 525 is not divisible by 2, since it does not end with 0, 2, 4, 6, or 8. 525/2 = 262.5
Yes, evenly
525 = 35 x 15
510
Yes.
There are eight pairs of numbers divisible by three that sum to 150, and 75 is also divisible by three, so: sum = 17 * 75 = 750 + 525 = 1275
No. To find out if a number is divisible by 3, just add up the numbers and if that number is divisible by three then the whole number is. So, 5+3+5=13, so it is not divisible by three. 5+2+5=12, so 525 IS divisible by three. Pretty neat, huh?!
The list is infinite. Here are some of them: 210, 315, 420, 525, 630, 735, 840.
510. To be divisible by 3, it must be a multiple of 3. Thus the required number is a multiple of both 3 and 5, which will be a multiple of their lcm: lcm(3, 5) = 15. → 495 ÷ 15 = 33 → first multiple of 15 greater than 495 is 15 x 34 = 510 → 525 ÷ 15 = 35 → last multiple of 5 less than 525 is 15 x 34 x 510 → number required is 510.
175