The valid factors of 76 are: 2,4,19,38 We are looking for a factor that leaves a remainder of 3 when divided by four. Since this implies that it is odd, it must be 19.
15
1098765433 will have a remainder of 1.
The remainder is always 31 or less.
0.0385
989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
24 is answer.
15
3
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
Nothing. The remainder has to be less than the divisor.
less than
1098765433 will have a remainder of 1.
The remainder is always 31 or less.
0.0385
It's 19 I'm awesome rite u evil people made me post then hav to solv my own question how could u
I would think 100 is the answer.
2.5