Question

OI 20. 7 Find the smallest number which when divided by 18. 12, and 24 leaves a remainder of 16, 10, and 22, respectively, 11​

Answer 1

Answer:

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Answer 2

hey there!!

Step-by-step explanation:

• The solution is 70.

• Let S be the desired smallest (non-negative) integer.

• Then there are integers a,b and c such that:

• S = 18a + 16 = 18(a+1) - 2 ---> S+2 = 18(a+1)

• S = 12b + 10 = 12(b+1) - 2 ---> S+2 = 12(b+1)

• S = 24c + 22 = 24(c+1) - 2 ---> S+2 = 24(c+1)

• So S + 2 is a multiple of each of 12, 18 and 24.

• Since the lowest common multiple of 12, 18 and 24 is 72,

we conclude that,

• S+2 = is a multiple of 72 and therefore

• S = 70 is the smallest candidate for a solution.

Checking, 70/18 = 3 rem 16, 70/12 = 5 rem 10 and 70/24 = 2 rem 22.

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