Find the smallest number that is divisible by 2, 3, 4, 5, 6 and add 1.
We need the least common multiple of 2, 3, 4, 5, 6.
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
LCM = product of common and not common prime factors with larger exponent.
LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60
To always have a remainder of 1, you need of add 1 to 60.
The number is 61.
Check:
61/2 = 30 remainder 1
61/3 = 20 remainder 1
61/4 = 15 remainder 1
61/5 = 12 remainder 1
61/6 = 10 remainder 1
Answer: C
Step-by-step explanation:
fcbgxhjkl;'
471/5=94.2
hope that helps!!
Answer:
The two linear equations represented in system A as :
3 x + 2 y =3 -------(1)
- 2 x - 8 y = -1 ------(2)
(1) × 2 + (2) × 3 gives
⇒ 6 x + 4 y - 6 x - 24 y = 6 -3
⇒ - 20 y = 3
⇒ y =
Putting the value of y in equation (1), we get
Two linear equation represented in system B is:
3. -x - 14 y =1
4. - 2 x - 8 y = -1
-2 ×Equation (3) + Equation (4)=
2 x +28 y- 2 x - 8 y= -2 -1
⇒ 20 y = -3
⇒y =
Putting the value of y in equation (3),we get
As Two system , that is system (A) and System (B) has same solution.
By looking at all the options , i found that Option (D) is correct. The two system will have the same solution because the first equation of System B is obtained by adding the first equation of System A to 2 times the second equation of System A.
