Even a set can be sincere, why not you?

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 a b \dfrac{a}{b} , where a a and b b are coprime positive integers, then find the value of a + b a+b .


The answer is 143.

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2 solutions

Kushal Bose
Nov 24, 2016

Let five consecutive integers are like a 2 , a 1 , a , a + 1 , a + 2 a-2,a-1,a,a+1,a+2 .Its sum is 5 a 5a

If 5 a 5a should ends with 5 5 then a a should be odd integer.

Here a a should starts with 3 3 and ends with 97 97 because minimum of a 2 = 1 i . e . a = 3 a-2=1\,\,i.e.\,\, a=3 and maximum of a + 2 = 99 i . e . a = 97 a+2=99\,\,i.e.\,\, a=97 .

So 3 a 97 3 \leq a \leq 97 .

Total number of set of five consecutive integers less than 100 is 95 95 because it starts with ( 1 , 2 , 3 , 4 , 5 ) (1,2,3,4,5) and ends with 95.96.97 , 98 , 99 ) 95.96.97,98,99) .

No of value of a = 48 a=48 and total number of sets of five consecutive integers is 95 95 .

Then probability is 48 95 \dfrac{48}{95}

Posted a similar problem .

@Nihar Mahajan

Harsh Shrivastava - 4 years, 6 months ago

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Nice problem! Will solve it later.

Nihar Mahajan - 4 years, 6 months ago

Exactly the intended solution. Here's an upvote from me!

Nihar Mahajan - 4 years, 6 months ago
Ashish Menon
Dec 20, 2016

Pick the maximum possible sum of the possible 5-tuples which is 95 + 96 + 97 + 98 + 99 = 485 95 + 96 + 97 + 98 + 99 = 485 .
Let x 2 , x 1 , x , x + 1 , x + 2 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 ) 485 x 97 \implies (x - 2) + (x - 1) + x + (x + 1) + (x + 2) \leq 485\\ \implies x \leq 97 .

First try ( x 2 ) + ( x 1 ) + x + ( x + 1 ) + ( x + 2 ) = 5 x = 1 ( x 2 ) + ( x 1 ) + x + ( x + 1 ) + ( x + 2 ) = 15 x = 3 ( x 2 ) + ( x 1 ) + x + ( x + 1 ) + ( x + 2 ) = 25 x = 5 ( x 2 ) + ( x 1 ) + x + ( x + 1 ) + ( x + 2 ) = 35 x = 7 (x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 5\\ x = 1\\ \\ (x - 2) + (x - 1) + x + (x+ 1) + (x + 2) = 15\\ x = 3\\ \\ (x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 25\\ x = 5\\ \\ (x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 35\\ \\ x = 7 .

So we observe the sets are sincere a 0 x \forall a0x \in AP 1 , 3 , 5 , 1,3,5, \cdots .

But when, x = 1 x = 1 , x 2 = 1 x - 2 = -1 and x 1 = 0 x - 1 = 0 which is not possible since x x is a positive inetegr.

So, the number of values of x x for which set is sincere is the number representing which term 97 is of the AP 3 , 5 , 7 , . . . 3,5,7,... which is 48 48 .

And the total number of 5-tubles consisting of consecutive positive integers less than 100 is 95 95 .

a b = 48 95 a + b = 48 + 95 = 143 \therefore \dfrac{a}{b} = \dfrac{48}{95}\\ \implies a + b = 48 + 95 = \boxed{143} .

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