Answer:
To determine how many 6-digit numbers of the form
568
568abc are divisible by 3, 4, and 5, we need to ensure the number satisfies the conditions for divisibility by these numbers.
Step 1: Divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Sum of digits in
568
568abc is
5
+
6
+
8
+
+
+
=
19
+
+
+
5+6+8+a+b+c=19+a+b+c.
To make
19
+
+
+
19+a+b+c divisible by 3,
+
+
a+b+c must be congruent to
2
(
m
o
d
3
)
2(mod3).
Step 2: Divisibility by 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Last two digits of
568
568abc are
bc.
Step 3: Divisibility by 5
A number is divisible by 5 if its last digit is 0 or 5.
Last digit of
568
568abc is
c.
Now, let's combine these conditions:
c must be 0 or 5 (divisible by 5).
bc must be divisible by 4.
Step 4: Counting possible values
Let's analyze the values of
+
+
a+b+c modulo 3:
If
+
+
≡
2
(
m
o
d
3
)
a+b+c≡2(mod3), possible values for
+
+
a+b+c are 2, 5, 8, 11, 14, 17.
For each valid sum
+
+
a+b+c, count how many pairs
(
,
)
(b,c) satisfy
bc being divisible by 4 and
c being 0 or 5:
If
=
0
c=0:
b can be 0, 2, 4, 6, 8 (5 choices).
If
=
5
c=5:
b can be 0, 2, 4, 6, 8 (5 choices).
So, for each valid
+
+
a+b+c, there are
5
+
5
=
10
5+5=10 combinations of
(
,
)
(b,c) that satisfy all conditions.
Step 5: Total number of 6-digit numbers
Calculate the total number of valid combinations for
,
,
a,b,c:
There are 6 possible sums
+
+
a+b+c that satisfy
+
+
≡
2
(
m
o
d
3
)
a+b+c≡2(mod3).
For each sum, there are 10 valid combinations of
(
,
)
(b,c).
Therefore, the total number of 6-digit numbers
568
568abc that are divisible by 3, 4, and 5 is
6
×
10
=
60
6×10=
60
.
