Misunderstood
A teacher has given the following equation to his four best students, and then asked how many integer answers it had.
[ x + 1 2 1 2 x + 1 ] = 1 0
Al considered his teacher's notation to be a mere parenthesis, and quickly shouted his answer. Bob considered [ x ] the floor function, and answered in time with his fellows Carl (which used the round function for his solution) and Dan (which preferred the ceiling function).
Which boy gave out the biggest answer?
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2 solutions
Very interesting question btw, I have never seen a question test this topic in such a way.
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Thank you Julian! I am very happy to contribute with new approaches on simple topics. This reminds of Mistakes Give Rise to Problems !
Feel free to share this question and use it anywhere, if you want to!
Should there be an equal sign with the inequality in Carl's solution near the 19/2? Wouldn't 9.5 round up to 10 and be correct from his point of view?
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Working with round ( 2 2 n + 1 ) has always been a problem for me. Fixed.
We can do it in a simpler way... To get the maximum no of solutions we must have the biggest range to operate the function. for Floor its from 10----- <11 for Round its from 9.5------<10.5 for Ceil its from 9------<10 All have a difference of 1,so we have to see the case in which the co-efficient of x gets the smallest value, so that the value of x gets the maximum, eg: 12x+1 =10 (x+12) gives 12x-10x=2x so its 119/2 if the co efficient is 1 in case of 11 it is 12x+1=11(x+12), gives 12x-11x=x so its 131/1 so we can infer from this the the lesser we get to the co-efficient of x, the bigger the value of x, hence the bigger the set....
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Shivam, can this be generalized for any random equation? Check
[ x + a a x + 1 ] = b
for integers ( a , b ) such that a > b + 1 .
PS: For a = b + 1 , can you find out Bob's answer? Note that this is independent of ( a , b ) !
2nd line 119/2<=X..........gives 59 <=X
Last line. 107/3 < X........ gives 35 <= X.
We may make it a little simple as follows . For floor function.
X
+
1
2
1
2
X
+
1
=
1
2
−
X
+
1
2
1
4
3
⟹
[
−
X
+
1
2
1
4
3
]
=
−
2
⟹
−
1
>
[
−
X
+
1
2
1
4
3
]
≥
−
2
Inequality is with -tive and reciprocal, thus the relation signs remain the same.and we have
1
/
1
>
1
4
3
X
+
1
2
≥
1
/
2
⟹
1
4
3
>
X
+
1
2
≥
1
4
3
/
2
1
3
1
>
X
≥
5
9
⟹
5
9
≤
X
≤
1
3
0
.
.
.
.
.
.
.
.
7
2
The rest also in the same way.
for Al ,no of solutions = 0 obvious. now, for the remaining three of them Let y = (12x+1)/(x+10)
condition required for bob 10<=y<11 condition required for dan 9<y<=10 condition required for carl 9.5<y<10.5
for all three of them ,change in y is 1 and as we know y approaches 12 when x->infinity
So integer solution for ben will be most .
Bob is very angry that you have mistook his name (who is Ben?), but well, my question sure is about misunderstandings!
Al's Solution: 1 2 x + 1 = 1 0 x + 1 2 0 ⇔ x = 2 1 1 9 ∈ Z . Al's answer was 0 (zero) .
Bob's Solution: 1 0 ≤ x + 1 2 1 2 x + 1 < 1 1 ⇔ 2 1 1 9 ≤ x < 1 3 1 ⇒ 6 0 ≤ x ≤ 1 3 0 .
Bob's answer was 7 1 .
Carl's Solution: 2 1 9 ≤ x + 1 2 1 2 x + 1 ≤ 2 2 1 ⇔ 5 2 2 6 ≤ x ≤ 3 2 5 0 ⇒ 4 6 ≤ x ≤ 8 3 .
Carl's answer was 38 .
Dan's Solution: 9 < x + 1 2 1 2 x + 1 ≤ 1 0 ⇔ 3 1 0 7 < x ≤ 2 1 1 9 ⇒ 3 6 ≤ x ≤ 5 9 .
Dan's answer was 24 .