Which is the smallest number that can be divided by 15, 16 and 21,leaving a remainder of 3?
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Answer:
The smallest number will of course be seven; however, you most likely meant the smallest number other than seven.
To meet the criteria stated, n-7 must be a multiple of 15, 21, and 18. Since 0 is a multiple of all numbers, 7 works for this, but to find the second-smallest, we will have to find the smallest number (othed than 0) that is a multiple of all three, the LCM.
We start with the prime factorization of each numbers
15=3×5
21=3×7
18=2×3×3
We now multiply the prime factors together, but factors that multiple numbers have in common ard only counted once. Therefore, the LCM is 2×3×3×5×7. Not that we still multiply by 3 twice because the number 18 has three as a factor twice. However, we don't count it a third our fourth time even though it is a factor of the other numbers. 2×3×3×5×7=630, so 630 is the LCM of 15, 21, and 18. Therefore 637 will have a reminder of 7 when divided by either of them.
Step-by-step explanation:
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I think it is 7 hope it help