Question

Find the smallest number which when divided by 21, 28, 36 and 45 leaves a
remainder 9 in each case.
in ounctly divisible hy 12. 15.​

Answer 1

Answer:

1269

Step-by-step explanation:

Let X be the smallest number with the desired properties.

Thus X−9 is the smallest integer divisible by 21,28 36,45

Which implies that X−9 is the lcm of 21,28,36 and 45.

LCM have a distributive property

LCM(21,28,36,45)=LCM(LCM(21,28),LCM(36,45))=LCM(84,180)=1260

Thus

X=1269

Answer 2

Answer:

1269

Step-by-step explanation:

Prime factors of 21,28,36 and 45

21=3*7

28=2*2*7

36=2*2*3*3

45=3*3*5

Thus LCM=2²*3²*5*7

=4*9*5*7

=180*7

=1260

Thus 1260 is the smallest number divisible by 21, 28, 36 and 45

Now the remainder is in all three case=9

Thus required number=1260+9=1269