30/21 = 1 with a remainder of 9
71 ÷ 9 = 7 with remainder 8
It is divisible by 8 because 202008/8 = 25251 But divided by 9 it will leave a remainder ----------------------- To test if a number is divisible by 8 add the ones digit to twice the tens digit to 4 times the hundreds digit; if this sum is divisible by 8 then so is the original number. By repeating the test on the sum until a single digit remains, only if this single digit is 8 is the original number divisible by 8, otherwise it gives the remainder when divided by 8 (except if it is 9, in which case the remainder is 1 - the excess of 9 over 8). For 202008: ones_digit + 2 × tens_digit + 4 × hundreds_digit = 8 + 2 × 0 + 4 × 0 = 8; so 208008 is divisible by 8. To test if a number is divisible by 9, add up the digits of the number; if this sum is divisible by 9, then so is the original number. By repeating the test of the sum until a single digit remains, only if this single digit is 9 is the original number divisible by 9, otherwise it gives the remainder when the original number is divided by 9. (This single digit is also called the "digital root" of the number.) For 202008: 2 + 0 + 2 + 0 + 0 + 8 = 12; 1 + 2 = 3; so 202008 is not divisible by 9; it has a remainder of 3 when divided by 9.
Suppose you want to find the gcf of 5040 and 1274. Make a list or a "ladder" starting with 5040 and 1274. Each item after that is the remainder of dividing the two numbers above. 5040 1274 1218 (5040 divided by 1274 leaves a remainder of 1218) 56 (1274 divided by 1218 leaves a remainder of 56) 42 (1218 divided by 56 leaves a remainder of 42) 14 (56 divided by 42 leaves a remainder of 14) 0 (42 divided by 14 leaves a remainder of 0). When you reach zero, the number before it (in this case 14) is the gcf. The gcf of 5040 and 1274 is 14. In general: Let a and b be the two numbers. repeat while (b >0) Let c = the remainder of a divided by b Let a=b Let b=c The gcf is a.
Not if you divide correctly.
2519
9 divided by 5,310 with no remainder = 0.001694915254237288
The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9.
9.4444
9.7778
23.25
85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11
506
9
0.1184
94.2222
206.3333