Question

Find the smallest number which when divided by 30, 40 and 60 leaves the remainder 7 in each case.

Answer 1

Hi friend, Harish here.

Here is your answer:

To find,

The smallest number when divided by 30, 40 & 60. leaving the same remainder 7 in each case.

Solution:

Now let us find the LCM of these three numbers.

⇒ 30 = 2 × 3 × 5

⇒ 40 = 2³ × 5

⇒ 60 = 2² × 3 × 5.

The LCM = 2³ × 3 × 5 = 120.

Then, 120 is the number which is exactly divisible. 

Then, 120 + 7 = 127. This number leaves remainder 7 when divided by those 3 number.

Hence 127 is the smallest number that leaves 7 as remainder which divided by those numbers.
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Hope my answer is helpful to you.

Answer 2

HERE'S YOUR ANSWER ✅✅✅

HERE, WE HAVE TO FIND LCM OF THE 3 NO.S

STEP 1 : Prime factorization :-

- 30 = 2 x 3 x 5

- 40 = 2³ x 5

- 60 = 2² x 3 x 5  

The LCM is - 2² x 3 x 5

                   = 120

STEP 2 : Since the remainder should be 7, we have to simply add 7 to 120 (since 120 leaves remainder 0)

120 + 7 = 127

Hope Helps!!

---SciFi273