22480 = 281 x 5 x 16 x 1. And 16 is 2^4, so you can have any combination of these factors that 22480 is divisible by [2 x 2 x 5 = 20, for example].
5 is not divisible by 1260. 1260 is divisible by 5 because 1260 is a multiple of 5.
625 Is divisible by 5
No, only numbers that end in 5 or 0 are divisible by 5.
no because 5,15,25,35,45,55,65,75,85,95,105...are not divisible by 10.10,20,30,40,50,60,70,80,90,100 are divisible by 10 and 5
No; 1+7+3+5= 16 16 isn't divisible by 3.
if Divisible by 5 = ends in 0 or 5 if Palindrome = spelled the same forwards as backwards then number begins and ends with either 0 or 5 5_5, or 0_0 Sum is divisible by 8 5+5=10 16=divisible by 8 565
No, 80 is bigger than 5.
105
No. To be divisible by 6, a number must be even (divisible by 2) and divisible by 3. Although 3058 is even, it is not divisible by 3 since 3+0+5+8 = 16 which is not divisible by 3. Thus is it not divisible by 6.
99968
To determine what number 745 is divisible by, you need to check for divisibility rules of different numbers. 745 is not divisible by 2 because it is not even. It is not divisible by 3 because the sum of its digits (7+4+5 = 16) is not a multiple of 3. It is not divisible by 5 because it does not end in 0 or 5. However, 745 is divisible by 5 because the last digit is 5.
160 is divisible by all three numbers, 10, 2, and 5. 160 / 10 = 16 160 / 2 = 80 160 / 5 = 32
22480 = 281 x 5 x 16 x 1. And 16 is 2^4, so you can have any combination of these factors that 22480 is divisible by [2 x 2 x 5 = 20, for example].
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.
Nope. 16 is divisible by 1, 2, 4, 8, 16.
16, it is divisible by 2, 4,8 in addtion to itself and 1