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04 Oct 2022, 02:35
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (03:05) correct
59%
(03:00)
wrong
based on 110
sessions
History
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Time
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Hungry chipmunk notices that, along a straight path, there are walnuts placed at intervals of 1 meter. The chipmunk wants to gather all the walnuts in his burrow, which is exactly where the middle walnut is. The chipmunk starts gathering the walnuts from the leftmost walnut and carries only one walnut at a time. If after completing the task, the chipmunk covered a distance of 300 meters and collected odd number of walnuts, then how many walnuts did the chipmunk collect ?
A. 11
B. 13
C. 15
D. 23
E. 25
A. 11
B. 13
C. 15
D. 23
E. 25
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04 Oct 2022, 11:39
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Let's assume that there are n walnuts to the left of the burrow. These walnuts have been represented in red.
From the premise we know that there another n walnuts to the left of the burrow, these walnuts have been represented in green.
One walnut, which is exactly in middle, is represented in black.
So in total there are 2n + 1 walnuts.
We know that the chipmunk gathers the walnuts from the leftmost position and carries only one walnut at a time.
Inference - The chipmunk starts from position \(A_{n}\) travels to \(X\) , drops the walnut at X and returns to \(A_{n-1}\) position. The distance covered in this process -
\(A_{n}\) to X = n meters
X to \(A_{n-1}\) = n - 1 meters
For the second walnut, the chipmunk travels from \(A_{n-1}\) to X, drops the walnut at X and returns to \(A_{n-2}\), covering (n-1) meters in the onward direction and (n-2) meters in the return journey.
This process continues till all the walnuts have been collected.
Summary of distance travelled by the chipmunk in collecting walnuts from the left side of the burrow
\(A_{n}\) to X = n
X to \(A_{n-1}\) = n - 1
\(A_{n-1}\) to X = n - 1
X to \(A_{n-2}\) = n - 2
\(A_{n-2}\) to X = n - 2
.
.
X to \(A_{1}\) = n - 2
\(A_{1}\) to X = 1
Sum of all the distances = n + 2 * (n-1) + 2 (n-2) + .... 2(2) + 2(1)
= 2(1 + 2 + 3 + ... (n-1)) + n
= 2(\(\frac{n*(n-1)}{2}\)) + n
= \(n^2 \)
The story doesn't end here
To collect the walnuts on the right of burrow, the chipmunk will follow the same process. It will first travel to \(B_{1}\), collect the walnut travel to X, drop the walnut and travel to \(B_{1}\). It will repeat this process till all the walnuts have been collected.
Summary of distance travelled by the chipmunk in collecting walnuts from the right side of the burrow
X to \(B_{1}\) = 1
\(B_{1}\) to X = 1
X to \(B_{2}\) = 2
\(B_{2}\) to X = 2
.
.
X to \(B_{n}\) = n
\(B_{n}\) to X = n
Total distance travelled = 1 + 1 + 2 + 2 + ... n + n
= 2 ( 1 + 2 + 3 + . . . . . . n)
= 2 \((\frac{n(n+1)}{2}\))
= \(n^2 + n\)
Given:
Total Distance = Left Distance + Right Distance = 300
\( n^2 + n^2 + n\) = 300
Simplifying
n = 12
Total walnuts collected = 2 * 12 + 1 = 25
IMO E
The detailed working is shown in the image attached.
P.S - Please ignore the bad drawing. The chipmunk looks like a rat
From the premise we know that there another n walnuts to the left of the burrow, these walnuts have been represented in green.
One walnut, which is exactly in middle, is represented in black.
So in total there are 2n + 1 walnuts.
We know that the chipmunk gathers the walnuts from the leftmost position and carries only one walnut at a time.
Inference - The chipmunk starts from position \(A_{n}\) travels to \(X\) , drops the walnut at X and returns to \(A_{n-1}\) position. The distance covered in this process -
\(A_{n}\) to X = n meters
X to \(A_{n-1}\) = n - 1 meters
For the second walnut, the chipmunk travels from \(A_{n-1}\) to X, drops the walnut at X and returns to \(A_{n-2}\), covering (n-1) meters in the onward direction and (n-2) meters in the return journey.
This process continues till all the walnuts have been collected.
Summary of distance travelled by the chipmunk in collecting walnuts from the left side of the burrow
\(A_{n}\) to X = n
X to \(A_{n-1}\) = n - 1
\(A_{n-1}\) to X = n - 1
X to \(A_{n-2}\) = n - 2
\(A_{n-2}\) to X = n - 2
.
.
X to \(A_{1}\) = n - 2
\(A_{1}\) to X = 1
Sum of all the distances = n + 2 * (n-1) + 2 (n-2) + .... 2(2) + 2(1)
= 2(1 + 2 + 3 + ... (n-1)) + n
= 2(\(\frac{n*(n-1)}{2}\)) + n
= \(n^2 \)
The story doesn't end here
To collect the walnuts on the right of burrow, the chipmunk will follow the same process. It will first travel to \(B_{1}\), collect the walnut travel to X, drop the walnut and travel to \(B_{1}\). It will repeat this process till all the walnuts have been collected.
Summary of distance travelled by the chipmunk in collecting walnuts from the right side of the burrow
X to \(B_{1}\) = 1
\(B_{1}\) to X = 1
X to \(B_{2}\) = 2
\(B_{2}\) to X = 2
.
.
X to \(B_{n}\) = n
\(B_{n}\) to X = n
Total distance travelled = 1 + 1 + 2 + 2 + ... n + n
= 2 ( 1 + 2 + 3 + . . . . . . n)
= 2 \((\frac{n(n+1)}{2}\))
= \(n^2 + n\)
Given:
Total Distance = Left Distance + Right Distance = 300
\( n^2 + n^2 + n\) = 300
Simplifying
n = 12
Total walnuts collected = 2 * 12 + 1 = 25
IMO E
The detailed working is shown in the image attached.
P.S - Please ignore the bad drawing. The chipmunk looks like a rat
Attachments
Screenshot 2022-10-05 000814.png [ 160.56 KiB | Viewed 5199 times ]
02 Apr 2024, 08:08
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Official Solution:
Hungry chipmunk notices that, along a straight path, there are walnuts placed at intervals of 1 meter. The chipmunk wants to gather all the walnuts in its burrow, which is exactly where the middle walnut is. The chipmunk starts gathering the walnuts from the leftmost walnut and carries only one walnut at a time. If after completing the task, the chipmunk covered a distance of 300 meters and collected an odd number of walnuts, then how many walnuts did the chipmunk collect?
A. 11
B. 13
C. 15
D. 23
E. 25
Check the image below:
There are \(n\) walnuts to the left of the burrow, \(n\) walnuts to the right of the burrow, and 1 walnut exactly in the middle, at the burrow. Hence, the total number of walnuts is \(n + n + 1=2n + 1\).
The chipmunk starts where the leftmost walnut is, so the first leg of the travel covers \(n\) meters. After that, the chipmunk reaches the burrow. To collect the remaining \(n-1\) walnuts, placed to the left of the burrow, the chipmunk should travel \(n-1\) meters twice: to the walnut and back to the burrow; \(n-2\) meters twice, to the walnut and back to the burrow; and so on until the first walnut to the left, where it travels 1 meter to the walnut and 1 meter back to the burrow. Thus, the chipmunk covers a total of:
\(2*1 + 2*2 + 2*3 + ... + 2(n-2) + 2(n-1)=2(1+2+3+...+(n-2)+(n-1))\) meters.
The sum within the parentheses represents the sum of consecutive integers from 1 to \(n-1\), which equals to \(\frac{first + last}{2} * (number \ of \ terms)\). Hence, we get:
\(2(1+2+3+...+(n-2)+(n-1))= 2(\frac{1 + (n-1)}{2}*(n-1))=n(n-1)\)
Don't forget the \(n\) meters the chipmunk covered to collect the first walnut to get the total distance covered to collect all walnuts to the left:
\(n(n-1) + n = n^2\) meters.
Similarly, we can calculate the distance the chipmunk will cover to collect walnuts placed to the right of the burrow, with an expectation that the chipmunk has have to cover \(n\) meters twice, not once: to the rightmost walnut and back, making the distance covered to collect walnuts placed to the right equal to \(n^2 + n\) meters.
Thus, we have that \(n^2 + (n^2 + n) = 300\), which gives \(2n^2 +n- 300=0\). Solving gives \(n=12\). Therefore, the number of walnuts is \(2n+1=25\).
Answer: E
Hungry chipmunk notices that, along a straight path, there are walnuts placed at intervals of 1 meter. The chipmunk wants to gather all the walnuts in its burrow, which is exactly where the middle walnut is. The chipmunk starts gathering the walnuts from the leftmost walnut and carries only one walnut at a time. If after completing the task, the chipmunk covered a distance of 300 meters and collected an odd number of walnuts, then how many walnuts did the chipmunk collect?
A. 11
B. 13
C. 15
D. 23
E. 25
Check the image below:
There are \(n\) walnuts to the left of the burrow, \(n\) walnuts to the right of the burrow, and 1 walnut exactly in the middle, at the burrow. Hence, the total number of walnuts is \(n + n + 1=2n + 1\).
The chipmunk starts where the leftmost walnut is, so the first leg of the travel covers \(n\) meters. After that, the chipmunk reaches the burrow. To collect the remaining \(n-1\) walnuts, placed to the left of the burrow, the chipmunk should travel \(n-1\) meters twice: to the walnut and back to the burrow; \(n-2\) meters twice, to the walnut and back to the burrow; and so on until the first walnut to the left, where it travels 1 meter to the walnut and 1 meter back to the burrow. Thus, the chipmunk covers a total of:
\(2*1 + 2*2 + 2*3 + ... + 2(n-2) + 2(n-1)=2(1+2+3+...+(n-2)+(n-1))\) meters.
The sum within the parentheses represents the sum of consecutive integers from 1 to \(n-1\), which equals to \(\frac{first + last}{2} * (number \ of \ terms)\). Hence, we get:
\(2(1+2+3+...+(n-2)+(n-1))= 2(\frac{1 + (n-1)}{2}*(n-1))=n(n-1)\)
Don't forget the \(n\) meters the chipmunk covered to collect the first walnut to get the total distance covered to collect all walnuts to the left:
\(n(n-1) + n = n^2\) meters.
Similarly, we can calculate the distance the chipmunk will cover to collect walnuts placed to the right of the burrow, with an expectation that the chipmunk has have to cover \(n\) meters twice, not once: to the rightmost walnut and back, making the distance covered to collect walnuts placed to the right equal to \(n^2 + n\) meters.
Thus, we have that \(n^2 + (n^2 + n) = 300\), which gives \(2n^2 +n- 300=0\). Solving gives \(n=12\). Therefore, the number of walnuts is \(2n+1=25\).
Answer: E
