can someone please list all counting numbers which leave a remainder of 4

K

kumorifox

New member
6, 9 and 18.

If you have a remainder of 4, the number has to be greater than 4. Also, if 4 remains, the number must divide evenly in the number-4, and 22-4 = 18. Therefore, the numbers have to be factors of 18, greater than 4. The only numbers that satisfy this are 6, 9 and 18.
 
6, 9 and 18.

If you have a remainder of 4, the number has to be greater than 4. Also, if 4 remains, the number must divide evenly in the number-4, and 22-4 = 18. Therefore, the numbers have to be factors of 18, greater than 4. The only numbers that satisfy this are 6, 9 and 18.
 
Let N be any such counting number.

By the division algorithm,

22 = NQ + 4, where 4 < N

So, NQ = 18, 4 < N.

Answer:
All factors of 18 that are greater than 4 = { 6, 9, 18 }

For your notes:
The division algorithm states that given two integers a and d, with d 0

There exist unique integers q and r such that a = qd + r and 0 r < | d |, where | d | denotes the absolute value of d.

The integer

q is called the quotient
r is called the remainder
d is called the divisor
a is called the dividend
 
Let N be any such counting number.

By the division algorithm,

22 = NQ + 4, where 4 < N

So, NQ = 18, 4 < N.

Answer:
All factors of 18 that are greater than 4 = { 6, 9, 18 }

For your notes:
The division algorithm states that given two integers a and d, with d 0

There exist unique integers q and r such that a = qd + r and 0 r < | d |, where | d | denotes the absolute value of d.

The integer

q is called the quotient
r is called the remainder
d is called the divisor
a is called the dividend
 
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