Question
Find the smallest number by which 845 must be multiplied so
that the product is divisible by 30.Solution
ATQ,
Since, 845 = 5 Γ 13 Γ 13
To be divisible by β30β, a number must be divisible by 2, 3, and 5.
Here, 845 is divisible by 5 only. So, we need to multiply it by both 2 and 3, i.e., 2 Γ 3 = 6.
Therefore, required answer is 6.
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