find the least number which when divided by 5, 6 , 8 leaves remainder 1, 2, 3 respectively.?

Answer 1

No Idea

Tried to calculate it but got over 10,000 and got bored

Tried it suplimenting 7 for 8 and got the answer 206

Answer 2

dividing by 8 giving remainder 3, then the number must be odd.
and dividing by 5 giving remainder 1, then the unit digit must be 1, since 6 is even.but dividing by 6 must give an odd remainder, so there musta been typo there.

if the numbers are 5, 6, 7 and 1, 2, 3, then i'll start over.

dividing by 6 with remainder of 2, x (unknown number) must be even.
dividing by 5 with remainder of 1, x's ones digit must be 6, since 1 is odd, not even.
dividing by 7 with remainder 3, and the ones must be 6, then the number must be in the form of 70k-4.

returning to 6 and 2, [(70k-4)-2] = 70k-6 must be divisible with 6. since -6 is already multiple of 6, then 70k must also be a multiple. the only way is k being a multiple of 3. you want the least, so we take k=3.

answer
= 70k-4 when k=3
= 70*3 -4
= 206

Answer 3

24

Answer 4

Let x be the number.
Given:
5k+1 = 6m+2 = 8n + 3 = x, where k,m,n are integers.
m = (8n+1)/6
k = (6m+1)/5 = ((8n+1) +1)/5 = (8n + 2)/5
k = (8n + 2)/5 (from equating expressions in k and n)
Thus,
Find the smallest n such that k and m are integers.
For k to be integer, n = 1, 6, 11, ...
but for m to be integer, no such n because 8n+1 is odd and cannot be divisible by 6.

Seems like something amiss in the question...

Answer 5

Let n be the number

n = 6p + 2, then n must be even

But n = 8k + 3, then n must odd

Sorry, there is no solution, there is not an n that is even and odd at the same time. Are you sure that you typed things correctly?

Ana

Answer 6

11

Answer 7

16

Answer 8

There is none. 6n + 2 is even number while 8k +3 is odd.

Answer 9

why would I want to do this>