Question

Find the lowest natural number that when divided by 116,148, and 170 leaves a remainder of 8 in each case please give me the answer this is urgent​

Answer 1

Answer:

616428

Is the answer clear or I will send it again

Answer 2

Answer:

The lowest number that when divided by 116,148 and 170 leaves the remainder 8 = 364828

Step-by-step explanation:

To find,

The lowest natural number that when divided by 116,148, and 170 leaves a remainder of 8

Solution:

The lowest number which is exactly divisible by 116,148, and 170 is the LCM(116,148,170)

Hence the lowest number that when divided by 116,148 and 170 leaves the remainder 8 is LCM(116,148,170) + 8

To find LCM(116,148,170)

Prime factorization of 116 = 2 × 2× 29

Prime factorization of 148 = 2 × 2× 37

Prime factorization of 170 = 2 × 5× 17

The least common multiple of 116,148 and 170 = 2 × 2 × 5× 17 ×29×37

= 364820

The lowest number that when divided by 116,148 and 170 leaves remainder 8 = LCM(116,148,170) + 8 =  364820 + 8 = 364828

∴The lowest number that when divided by 116,148 and 170 leaves the remainder 8 = 364828

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