Question

plsee solve thisssss

Answer 1

Dividing by 35 divides by both 5 and seven. If the initial remainders were equal, then it'd be that, but clearly that is not the case.

Answer 2

Therefore, the number must be some multiple of 35 plus a certain number which has these properties. The smallest such number is equal to 5x+3 and 7y+2. Set those equations as equal: 5x+3=7y+2. Now, we see that 7y=5x+1. What multiple of seven is one over a multiple of five? (guess and check is fast enough)

Answer 3

ur approach is right just a little mistake:\[x=5k+3=7m+2\]not \(2m+7\)

Answer 4

plse can u solve it

Answer 5

ok, see\[5k+3=7m+2\]\[k=\frac{7m-1}{5}=m+\frac{2m-1}{5}\]in order for k to be an integer \(2m-1=5t\) and again\[m=\frac{5t+1}{2}=2t+\frac{t+1}{2}\]finally\[t=2s-1\]it gives\[m=2(2s-1)+s=5s-2\]\[x=7(5s-2)+2=35s-12\]

Answer 6

any integer

Answer 7

a new variable

Answer 8

yes

Answer 9

now, what after that
i mean how to solve next

Answer 10

we are done\[x=35s-12\]what will be the remainder if u divide it by 35?

Answer 11

i have divide 35s -12 by 35

Answer 12

yes

Answer 13

-12

Answer 14

exactly or -12+35=23

Answer 15

the remainder is 23 so :)

Answer 16

okay
@mukushla
thank u so much

Answer 17

very welcome

Answer 18

@mukushla
only one question , how s came and what is s called