Divisible by 5?
There is a math teacher in a school whose favorite number is a positive integer (let's call it x ) . She told her 5 students--Alice, Beatrice, Candice, Denise, Eunice--the number x , and the following conversation took place:
Alice:
"
x
+
1
is not divisible by 5."
Beatrice:
"
x
+
2
is not divisible by 5."
Candice:
"
x
+
3
is not divisible by 5."
Denise:
"
x
+
4
is not divisible by 5."
Eunice:
"
x
+
5
is not divisible by 5."
Without knowing x , can we guarantee that at least one of these 5 students is lying?
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3 solutions
Moderator note:
This solution makes nice use of visual intuition! Even with a pure number theory or algebra problem it can be possible to invoke visuals to aid in understanding.
We can think whether LaTeX: x is even or odd. Let's study the 2 cases for A, B, C, D and E..
1) Let LaTeX: x = 2 k ( k ∈ Z ) is even :
5 devides LaTeX: ( 2 k + 1 ) an LaTeX: odd number only if LateX: ( 2 k + 1 = 5 ) o r ( k = 2 )
5 devides LaTeX: ( 2 k + 2 ) an even number only if LateX: ( 2 k + 2 = 5 ) No solution.
5 devides LaTeX: ( 2 k + 3 ) an LaTeX: odd number only if LateX: ( 2 k + 3 = 5 ) o r ( k = 1 )
5 devides LaTeX: ( 2 k + 4 ) an even number only if LateX: ( 2 k + 4 = 5 ) No solution.
5 devides LaTeX: ( 2 k + 5 ) an LaTeX: odd number only if LateX: ( 2 k + 5 = 5 ) o r ( k = 0 )
2) Let LaTeX: x = 2 k + 1 ( k ∈ Z ) is odd :
5 devides LaTeX: ( 2 k + 2 ) an even number only if LateX: ( 2 k + 2 = 5 ) No solution.
5 devides LaTeX: ( 2 k + 3 ) an LaTex: odd number only if LateX: ( 2 k + 3 = 5 ) o r ( k = 1 )
5 devides LaTeX: ( 2 k + 4 ) an even number only if LateX: ( 2 k + 4 = 5 ) No solution.
5 devides LaTeX: ( 2 k + 5 ) an LaTeX: odd number only if LateX: ( 2 k + 5 = 5 ) o r ( k = 0 ) .
5 devides LaTeX: ( 2 k + 6 ) an even number only if LateX: ( 2 k + 6 = 5 ) No solution.
To sumerise, in the first case : LaTeX: Alice, Candice or Eunice lie. >In the second case, LaTeX: Beatrice or Denise lie.
Nice explanation! Visualizing the problem on a number line definitely helps :)
Right. Another way to think about this is:
Suppose that they are all telling the truth, then the product ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) is also not a multiple of 5. But expanding this expression gives x ( ⋯ ) + 5 ! = x ( ⋯ ) + 5 ( 2 4 ) , which is the sum of two integers that are divisible by 5, so this product is also divisible by 5, which is absurd.
X=-2... Then sequence is -1, 0, 1, 2, 3... None of them are divisible by 5
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0 is divisible by 5, because 0 = 5 x 0.
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Also, the problem stated a positive integer.
If 0 = 5 x 0 then 5 / 0 = 0 Isn't it ?
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@Siddharth Rai – No, this is not correct. Zero does not have a multiplicative inverse, so division by zero (i.e. multiplying by its inverse) is not defined.
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@Pranshu Gaba – Correct... Thanks for explaining
The problem states x is a positive integer.
I think it is not that "at least one of the students is lying" but that "only one of the students is lying". What do you think?
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Both are correct. Note that if x = 1 is true, then so is x ≥ 1 .
what if it is a decimal
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The question says that x is a positive integer .
If we took the pigeon hole principal method, there are one number which is divisible by 5 from five consecutive number....
I don't see the connection. How is pigeonhole principle related here?
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If we divide a number by 5, the remainder of some consecutive number can 0, 1, 2, 3, 4, 0, 1........... So there are maximum 4 consecutive that are not divided by 5. If we took 5 as pigeon and 4 as pigeonhole so there should be two number which remainder is same. But there are no consecutive number which remainder are same. So there should be a number which is divide by 5.
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There will always be a multiple of 5 between an unknown integer (x) and x add five. Let's suppose x was equal to 13. X add 2 is divisible by 5.
Let's take any number. For ex - 1 and do the tasks. 1+1 = 2 // 1+2 = 3 // 2+3 =5 // Woahh! We got a number divisible by 5, so no matter what integer it is, if u do all these stuff u get it divisible by 5
If the statements made by the five students are true, there are five consecutive integers, none of which are a multiple of five. The following argument shows that this is impossible.
Imagine the integer number line stretched out with the the multiples of five painted red. Between each red number there are four integers. So there can be at most four consecutive integers which are not multiples of five. So at least one of the students is lying.