No.50 multiplied by 3 would give you the answer of 150 remainder 1.So 151 is not divisible by 3.
151 is prime, and is thus only divisible by 1 and itself.
Yes, not evenly, quotient is 75, remainder is 1
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
Well, honey, anything that is a multiple of 453 is divisible by 453. So, if you find a number that can be evenly divided by 453 without leaving a remainder, then congratulations, you've got yourself a winner. Just remember, math doesn't have to be a drag, so go ahead and divide to your heart's content.
All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9. However, all numbers divisible by 9 are also divisible by 3 because 9 is divisible by 3.
Yes. The result is 151
There are 151 3-digit numbers that are divisible by 6.
151 is prime, and is thus only divisible by 1 and itself.
Yes, not evenly, quotient is 37, remainder is 3
No. 302 is divisible by 1, 2, 151, 302.
No. 755 is divisible by 1, 5, 151, 755.
151 is prime. It is only evenly divisible by itself and one.
1, 23, 151, 3473
no, but it is divisible by: 1, 13, 151, 1963, 4153, 53989, 627103, 8152339
1, 2, 3, 4, 6, 9, 12, 18, 36, 151, 302, 453, 604, 906, 1359, 1812, 2718, 5436
Yes, not evenly, quotient is 75, remainder is 1
By 3 only.It is not divisible by 2, 4, 5, 6, 9 & 10.Division tests:2:If the number is even, ie if the last digit is even (0, 2, 4, 6, 8) it is divisible by 2. 633 is odd (last digit is 3 which is not an even digit), so 633 is not divisible by 2.3:Sum the digits; if the sum is divisible by 3, the original number is divisible by 3. (Can repeat the summing until a single digit remains; if this digit is 3, 6 or 9 (ie divisible by 3) then the original number is divisible by 3.)6 + 6 + 3 = 151 + 5 = 66 is divisible by 3, so 663 is divisible by 3.4:Add the last (units) digit to twice the previous (tens) digit; if this sum is divisible by 4, so is the original number. (Can repeat summing until a single digit remains; if this digit is 4 or 8 (ie divisible by 4) then the original number is divisible by 4.)6 x 2 + 3 = 151 x 2 + 5 = 77 is not divisible by 4, so 663 is not divisible by 4.Note: all multiples of 4 are even; 633 is odd so it cannot be divisible by 4 (the above test does not need to be done).5:The last digit of the number must be 0 or 5. 3 is not 0 nor 5, so 663 is not divisible by 5.6:Number must pass the 2 and 3 tests (above). Fails 2 test (above), so 663 is not divisible by 6.9:Sum the digits; if the sum is divisible by 9, the original number is divisible by 9 (Can repeat the summing until a single digit remains; if this digit is 9 (ie divisible by 9) then the original number is divisible by 9)6 + 6 + 3 = 151 + 5 = 66 is not 9, so 663 is not divisible by 9.10:Last digit must be 0. 3 is not 0, so 663 is not divisible by10.