Question

Q3:-Complete and write the following activity: Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.​

Answer 1

Answer:

365 is the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

Step-by-step explanation:

We must find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

Let the number be N

Then, if we subtract 5 from N then, that number must be divisible by 8, 9, 10, 15 and 20.

That is, when the remainder is subtracted from the dividend, it becomes divisible by the divisor right!!!

Thus, we must find the lowest common multiple that can be divisible by 8, 9, 10, 15 and 20

So,

LCM of 8, 9, 10, 15 and 20

8 = 2³

9 = 3²

10 = 2 × 5

15 = 3 × 5

20 = 2² × 5

Then, LCM will be the product of primes with the highest power.

So,

LCM = 2³ × 3² × 5

= 8 × 9 × 5

= 360

Now,

N - 5 = 360

N = 360 + 5

N = 365

That is 360 is exactly divisible by 8, 9, 10, 15 and 20, but we must also get a remainder of 5 so we must add 5 to the LCM.

Hence,

365 is the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

(In short, for your exams, take the LCM of the given numbers which has to be divided and then add the Remainder that will give you the answer.)

Hope it helped you and believing you understood it...All the best