Math, asked by rudragiri3908, 1 year ago

What is the least number which when we divided 8 12 and 16 leaves 3 the remainder each case but when divided by 7 leaves no remainder?

Answers

Answered by REDRAGON

To find the Least number X,

X should leave a remainder 3 when divided by 16 , 12, 8.

X Should completely divisible by 7.

So, Let's start with 16 because we can reach X quickly [ 16 is big number and we are moving 16 steps everytime]

Given condition,

X % 16 = 3

If a number X when divided by 16 gives some remainder Y then same number when divided by 8 gives the same remainder Y because 8 is a multiple of 16.

So need not bother about 8.

Now coming to number 12 with 16,

3 * 2 * 2 = 12 vs 2 * 2 * 2 * 2 = 16

12 is a contains multiple 3 where 16 does not contain the same.

So X-3 should be a multiple of 3.

Now X can be

51(48+3), 99(96+3), 147(144+3)…… (Choose multiples of 48 (16*3))

Now let’s search for numbers divisible by 7 in the above list.

147 is one such number and least number.

Answer = 147

Alternate Method :

Let us find the LCM of

8 = 2x2x2

12 = 2x2x3

16 = 2x2x2x2

LCM = 2x2x2x2x3 =48.

Let the number be (48n+3)/7 = Q, or

7Q = 48n + 3, or

7Q-3 = 48n

Check: 147 is divisible by 7. 147 divided by 8, 12 and 16 leaves a remainder of 3. Correct.

Hope This Helps :)

Answered by Anonymous

Answer: