Find the least number which when divided by 12 and 16 leaves the reminder 3 and 7 respectively. Plzz help in this question ​

Explanation:

Let n be the number we want to find. We want n to be as small as possible, but also be a positive integer. Intuitively, we can see that n cannot be smaller than 16; otherwise, we don't meet the remainder requirements.

Divide n over 12 and we get some quotient x and remainder 3

So,

n/12 = (quotient) + (remainder)/12

n/12 = x + 3/12

Multiply both sides by 12 to end up with

n = 12x + 3

Similarly, if we divide over 16 we get some other quotient y and remainder 7

n/16 = y + 7/16

which turns into

n = 16y + 7

after multiplying both sides by 16

We have these two equations

Apply substitution and do a bit of rearranging like so

12x+3 = 16y+7

12x-16y = 7-3

4(3x-4y) = 4

3x-4y = 4/4

3x-4y = 1

The goal from here is to find the smallest positive integers x and y that make that equation true. We have a few options here and they are

Whichever method you use, you should find that (x,y) = (3,2) is the point we want.

Note how:

3x-4y = 1

3(3)-4(2) = 1

9-8 = 1

1 = 1

So that verifies (3,2) is on the line 3x-4y = 1.

Because x = 3 and y = 2, we know that

n = 12x + 3

n = 12*3 + 3

n = 39

and we can see that

n = 16y + 7

n = 16*2 + 7

n = 39

So 39 is the smallest such integer such that when we divide it over 12 and 16, we get remainders 3 and 7 respectively.

Here's a quick verification that we've fit the requirements.

39/12 = 3 remainder 3

39/16 = 2 remainder 7

We know we hit the smallest value of n because (x,y) was made to be the smallest positive integer solution to 3x-4y = 1. There are infinitely many positive integer (x,y) solutions to 3x-4y = 1, which in turn means there are infinitely many numbers n that satisfy the remainder conditions (but n is not the smallest possible in those cases).

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24

answer:   y=2x-3

step-by-step explanation: