There exists infinitely many solutions to the problem.
Let the required number be x:
By Euclid's Division Lemma we have
x = 9q + 1(Since dividend is 9 and remainder is 1)
Put any value of q(q must be natural number), say 5.
On putting q = 5, we get x = 9*5 + 1 = 46.
Put any value of q and you will get different value of x.
5 will go into 9 once with a remainder of 4.
q=-1 You do the opposite operation to get rid of a number next to a variable. In this case, since you are multiplying you divide the other side with that number. So you take 9 and divide -9 by 9. This gets rid of the 9 next to the q, and you are left with -1 = q.
3, looks as if you add 9 to the first number, then add 1, divide by 5, add 9, add 1, divide by 5 to get 3
370740864321110.1
The remainder can be any number between 1 and 8 .
49.1481
5 will go into 9 once with a remainder of 4.
A number divided by 2 its reminder is 1, A number divided by 3 its reminder is 2, A number divided by 4 its reminder is 3, A number divided by 5 its reminder is 4, A number divided by 6 its reminder is 5, A number divided by 7 its reminder is 6, A number divided by 8 its reminder is 7, A number divided by 9 its reminder is 8, A number divided by 10 its reminder is 9 what is that number?
30/21 = 1 with a remainder of 9
9 is the largest integer that can divide into both 36 and 81 evenly with no remainder.
It is 9 as 9 /1 is 9 divide by 1 and 1 goes in to 9 nine times.
The number is 72 because (72/9)-7 = 1
The smallest positive whole number that will evenly divide into 9 and 19 is 1. This is true for any number. All numbers must divide by 1!
It is 2520.
It goes: 28/3 = 9 times with a remainder of 1
q=-1 You do the opposite operation to get rid of a number next to a variable. In this case, since you are multiplying you divide the other side with that number. So you take 9 and divide -9 by 9. This gets rid of the 9 next to the q, and you are left with -1 = q.
3, looks as if you add 9 to the first number, then add 1, divide by 5, add 9, add 1, divide by 5 to get 3