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Is 745 divisible by 2 3 5 9 or 10?
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.
Is 710 divisible by 2 3 5 9 or 10?
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
Is 55557 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
Is the number 75 divisible by 2 3 4 5 9 10?
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
How do you know if the number is divisible by 15?
All numbers divisible by 5 (of which 15 is a multiple) have a final digit of 0 or 5. All numbers divisible by 3 (of which 15 is a multiple) have the sum of the digits totalling 3 or a multiple of 3. Therefore, a number is divisible by 15 if the sum of its digits total 3 or a multiple of 3 and its final digit is 0 or 5. Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by 3. Final digit 5. This number is divisible by 15. (32085 ÷ 15 = 2139) 7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by 15.
Is 745 divisible by 2 3 5 9 or 10?
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.
What is the divisibility rules of 105?
For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.If the final digit of the number is 0 or 5, the original number is divisible by 5.Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.
Is 710 divisible by 2 3 5 9 or 10?
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
Is 55557 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
How many 3 digit numbers are divisible by 7?
The smallest 3-digit multiple of 7 is 105 = 15*7 The largest 3-digit multiple of 7 is 994 = 142*7 So there are 142-14 = 128 3-digit multiples of 7, ie 128 3-digit numbers that are divisible by 7.
What is a divisibility test for 7?
for 2 digits... multiply the last digit by 5, add the result to the first digit... example... 63.. (3)(5)=15 15+6=21 21 is divisible by 7, so 63 is divisible by 7...
The number 630 is divisible by what single - digit numbers?
630 is divisible by...2 or 3 or 5 or 6 or 7 or 9. your welcome
Is 5715 divisible by 2 3 5 9 or 10?
5715 is not divisible by 2: the last digit is not an even number or 05715 is divisible by 3: as the last two digits when added together is a multiple of 35715 is divisible by 5: as the last digit is a 5 or 05715 is divisible by 9: as if u added all the digits of the number it will be divisible by 9, in this case 5 +7+1+5=18 and 18 is a multiple of 95715 is not divisible by 10: as the last digit is not a 0
What is a 3 digit number divisible by 7 but not 2 the sum of your digits is 4?
I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I
Is the number 75 divisible by 2 3 4 5 9 10?
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
What are the single-digits divisible by the number 630?
There are no single-digit numbers divisible by 630. 630 is divisible by 1, 2, 3, 5, 6, 7 and 9.
What 4 digit number that is divisible by 5 and 7?
5678 \
