find the least no which when divided by 17, 16 ,15 leaves the remainder 3, 10, 0 respectively u have to explain the step

Find the least number such that the
remainder when divided by 17 is 3, when divided by 16 is 10, and when divided by 15 is
0.

If I have the question correctly stated,
then the answer from your book is wrong since 3910/15 is not a whole
number.

(1) The number must be a multiple of 15 since the
remainder is 0.

(2) We look for the first multiple of 15
that is 10 more than a multiple of 16. We find that 90=15*6 and 16*5+10=90. The LCM of
15 and 16 is 240, so the answer will be a number of the form
90+240x.

(3) We need a number of the form 90+240x that is 3
more than a multiple of 17. We notice that 90=17*5+5, 330=17*19+7 so we try the other
direction and see that -150=17*(-9)+3. We presumably want the smallest positive number
or the problem is unanswerable, so we add the LCM of 240 and 17 which is 4080 to get
3930.

Note that 3930=15*262, 3930=16*245+10, and
3930=17*231+3 so 3930 is the
solution.

If you are studying
modular arithmetic, note that the number x has the following
properties:

x=0(mod15)
x=3(mod17)
x=10(mod16)

So
we are looking for a w such that

15w=10(mod16) or
w=6(mod16)
15w=3(mod17)   or 5w=1(mod17)
Solving for w yields w=262
and 15(262)=3930, the number we are seeking.