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Q: What numbers are divisible to 745?

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If the last digit is 0 or 5 then the number is divisible by 5. Last digit of 745 is 5 so, 745 is divisible by 5.

1, 5, 149, 745.

1, 3, 5, 15, 149, 447, 745, 2235

By any of its factors which are: 1, 5, 149 and 745

Out of that list, just 5. ------------------------------------------------ To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8} The last digit is 5 which is not even (it is odd), so 745 is not divisible by 2 To be divisible by 3 the sum of the digits must also be divisible by 3; if the summing is repeated until a single digit remains, then the original number is only divisible by 3 if this single digit is divisible by 3, ie it is one of {3, 6, 9} 745 → 7 + 4 + 5 = 16 16 + 1 + 6 = 7 7 is not divisible by 3 (it is not one of {3, 6, 9}), so 745 is not divisible by 3 To be divisible by 5, the last digit must be 0 or 5 The last digit of 745 is 5 which is one of {0, 5}, so 745 is divisible by 5 To be divisible by 9 the sum of the digits must also be divisible by 9; if the summing is repeated until a single digit remains, then the original number is only divisible by 9 if this single digit is 9 (otherwise this single digit gives the remainder when the original number is divided by 9) 745 + 7 + 4 + 5 = 16 16 → 1 + 6 = 7 7 is not 9, so 745 is not divisible by 9 (the remainder is 7) To be divisible by 10 the last digit must be 0 The last digit of 745 is 5 which is not 0, so 745 is not divisible by 10. 745 is not divisible by 2, 3, 9, 10 745 is divisible by 5.

No, it is only divisible by 1, 5, 25, 149, 745, 3725.

1, 5, 25, 149, 745, 3725

All numbers are divisible by 1.

147 + 148 + 149 + 150 + 151 = 745

Numbers are divisible by their factors. Multiples are divisible by their numbers.

Composite numbers are divisible by prime numbers.

Prime numbers are divisible because any numbers that are divisible are prime. If a number isn't divisible, it isn't prime. Prime numbers have to be divisible by at least one pair of numbers to be prime.

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