Competive Exam Math 8 : How many numbers up to (?) are divisible by both (?) and (?).
Q. How many numbers up to 550 are divisible by both 4 and 5 ? Q. 550 āĻ˛ā§āĻā§ āĻāĻŋāĻŽāĻžāĻ¨āĻāĻž āĻ¸āĻāĻā§āĻ¯āĻž 4 āĻā§°ā§ 5 āĻĻā§āĻ¯āĻŧā§āĻāĻžā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āĻ¯ ?
Trick | āĻā§ā§°āĻŋāĻ
A number divisible by both 4 and 5 must be divisible by their LCM.
4 and 5 both divisible (4 āĻā§°ā§ 5 āĻĻā§āĻ¯āĻŧā§āĻāĻžā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āĻ¯ ) ⇒ LCM(4, 5) = 20
Now,
550 ÷ 20 = 27 remainder 10
So, there are 27 multiples of 20 up to 550.
āĻ ā§°ā§āĻĨāĻžā§, 550 āĻ˛ā§āĻā§ 20 ā§° 27āĻāĻž āĻā§āĻŖāĻŋāĻ¤āĻ āĻāĻā§āĨ¤
Formula | āĻ¸ā§āĻ¤ā§ā§°
Count = ⌊550 / 20⌋ = 27
Ans | āĻāĻ¤ā§āĻ¤ā§°: 27
Q. How many numbers up to 666 are divisible by both 6 and 10 ? Q. 666 āĻ˛ā§āĻā§ āĻāĻŋāĻŽāĻžāĻ¨āĻāĻž āĻ¸āĻāĻā§āĻ¯āĻž 6 āĻā§°ā§ 10 āĻĻā§āĻ¯āĻŧā§āĻāĻžā§°ā§ āĻŦāĻŋāĻāĻžāĻā§āĻ¯ ?
Option: A. 11, B. 22, C. 21, D.23
Trick | āĻā§ā§°āĻŋāĻ
First, find the LCM of 6 and 10.
LCM(6, 10) = 30 (6 āĻā§°ā§ 10 ā§° LCM = 30)
Now divide: 666 ÷ 30 = 22 remainder 6
So, there are 22 multiples of 30 up to 666. (āĻ ā§°ā§āĻĨāĻžā§, 666 āĻ˛ā§āĻā§ 30 ā§° 22āĻāĻž āĻā§āĻŖāĻŋāĻ¤āĻ āĻāĻā§āĨ¤)
Formula | āĻ¸āĻšāĻ āĻ¸ā§āĻ¤ā§ā§°
Count = ⌊666/30⌋ = 22
Ans: B. 2