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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.
Check your divisibility rules: 2: One's digit is an even number YES 3: The sum of the digits is divisible by 3 NO: 7 + 1 + 0 = 8 which is not divisible by 3 5: One's digit is a 5 or a 0 YES 9: The sum of the digits is divisible by 9 NO: 7 + 1 + 0 = 8 which is not divisible by 9 10: One's digit is a 0 YES
No, 1017 is not divisible by 5. A number is divisible by 5 only if its last digit is 0 or 5. Since the last digit of 1017 is 7, it is not divisible by 5
No, 55557 is not divisible by 5. To be divisible by 5, the last digit must be 0 or 5. The last digit of 55557 is 7, and 7 is neither 0 nor 5, thus 55557 is not divisible by 5
To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit of 75 is 5, which is not one of these so it is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is one of {3, 6,9} so it is divisible by 3. To be divisible by 4, add the last (ones) digit to twice the previous (tens) digit; if this sum is divisible by 4, then so is the original number. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {4, 8} is the original number divisible by 4. For this gives: 75→5 + 2×7 = 19 19→9 + 2×1 = 11 11→1 + 2×1 = 3 3 is not one of {4, 8} so it is not divisible by 4. To be divisible by 5, the last digit must be one of {0, 5}. The last digit of is 5 which is one of {0, 5} so it is divisible by 5. To be divisible by 9, sum the digits of the number and if this sum is divisible by 9, then the original number is divisible by 9. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is 9 is the original number divisible by 9. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is not 9 so it is not divisible by 9. To be divisible by 10, the last digit must be 0 The last digit is 5 which is not 0, so it is not divisible by 10. → 75 is divisible by 3 and 5 75 is not divisible by 2, 4, 9, 10