Math, asked by hardikbmali24, 2 months ago

Which is the smallest number that can be divided by 15, 16 and 21,leaving a remainder of 3?

You Have any opsion or not?

Answers

Answered by abhishek7528
0

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:

pls mark on brainlist

Answered by naginpatel8866013809
0

I think it is 7 hope it help

Similar questions