Yes - 2418/6 = 403
403 is divisible by 1, 13, 31 and 403.
403 is divisible by no number
No (not if you want a whole number as an answer).
403÷8 gives 50 as quotient and 3 as remainder. Dividend- remainder=divisor ×quotient 403-3=8*50 which is 400. our value is 403 So increase divisor 8*51=408. 403+5 gives 408. So 5 must be added to 403 to get a no divisible by 8.
67.1667
67.1667
6 is not divisible by 162. 162 is divisible by 6.
2418
0.0149
403 minutes = 6 hours and 43 minutes
806 would have to be divisible by 3 and 2 (6 =3*2). OK, let us see. The last digit is even, so sure, 806 is divisible by 2. But adding the three digits together, I get 14 (=8+0+6), which is not divisible by 3. So, no, 806 is not divisible by 6. I have described the fun way. The more direct way would be doing factorization (806 = 403*2 = ? Uh oh, 403 seems to be a prime factor, because it is indivisible by 2, 3, 5, 7, 11, and ...; no, wait, it is divisible by 13. So I have 806 = 13 * 31 * 2.) or using a calculator. The rules that I can remember are as follows. divisible by 2, if the number is even; 3, if the sum of the digits is divisible by 3; 5, if the last digit is 0 or 5; 11, if the sum of odd digits equals the sum of even digits (e.g. 121 is divisible by 11 because 1+1 = 2); the rule for 7 is a little difficult to remember, so I consult .mathsisfun.com/divisibility-rules.html.