Question

Find the smallest number which leaves the remainders 13,41 and 29 at the end when divided by 20,48 and 36 respectively

Answer 1

Answer:

First find out the L.C.M. of 13,41,29 then add the answer into 28,48,36.

Answer 2

Answer: 713

Step-by-step explanation:

• Given:- 20, 48 and 36 when divides a number leaves remainder 13, 41, and 29 when divides a number.
• To find:- The smallest required number
• Solution:-

       LCM of 20, 48 and 36.

       20=2×2×5

       48=2×2×2×2×3

       36=2×2×3×3

       LCM=2×2×3×3×2×5

               = 720

We subtract 7 from the LCM and we get 713 which leaves remainder 13,41 and 29 when divided by 20,48 and 36 respectively.

• Hence, the correct answer is 713.

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