Calvin calls a set of any five consecutive positive integers (each less than 100) as sincere if their sum ends with the digit 5.
If the probability that a given set of any five consecutive positive integers (each less than 100) is sincere can be expressed as b a , where a and b are coprime positive integers, then find the value of a + b .
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Pick the maximum possible sum of the possible 5-tuples which is
9
5
+
9
6
+
9
7
+
9
8
+
9
9
=
4
8
5
.
Let
x
−
2
,
x
−
1
,
x
,
x
+
1
,
x
+
2
be a group of sincere positive integers.
⟹
(
x
−
2
)
+
(
x
−
1
)
+
x
+
(
x
+
1
)
+
(
x
+
2
)
≤
4
8
5
⟹
x
≤
9
7
.
First try ( x − 2 ) + ( x − 1 ) + x + ( x + 1 ) + ( x + 2 ) = 5 x = 1 ( x − 2 ) + ( x − 1 ) + x + ( x + 1 ) + ( x + 2 ) = 1 5 x = 3 ( x − 2 ) + ( x − 1 ) + x + ( x + 1 ) + ( x + 2 ) = 2 5 x = 5 ( x − 2 ) + ( x − 1 ) + x + ( x + 1 ) + ( x + 2 ) = 3 5 x = 7 .
So we observe the sets are sincere ∀ a 0 x ∈ AP 1 , 3 , 5 , ⋯ .
But when, x = 1 , x − 2 = − 1 and x − 1 = 0 which is not possible since x is a positive inetegr.
So, the number of values of x for which set is sincere is the number representing which term 97 is of the AP 3 , 5 , 7 , . . . which is 4 8 .
And the total number of 5-tubles consisting of consecutive positive integers less than 100 is 9 5 .
∴ b a = 9 5 4 8 ⟹ a + b = 4 8 + 9 5 = 1 4 3 .
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Let five consecutive integers are like a − 2 , a − 1 , a , a + 1 , a + 2 .Its sum is 5 a
If 5 a should ends with 5 then a should be odd integer.
Here a should starts with 3 and ends with 9 7 because minimum of a − 2 = 1 i . e . a = 3 and maximum of a + 2 = 9 9 i . e . a = 9 7 .
So 3 ≤ a ≤ 9 7 .
Total number of set of five consecutive integers less than 100 is 9 5 because it starts with ( 1 , 2 , 3 , 4 , 5 ) and ends with 9 5 . 9 6 . 9 7 , 9 8 , 9 9 ) .
No of value of a = 4 8 and total number of sets of five consecutive integers is 9 5 .
Then probability is 9 5 4 8