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10 Jun 2022, 23:45
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C
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Difficulty:
55%
(hard)
Question Stats:
38% (02:29) correct
63%
(03:09)
wrong
based on 8
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Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number 1, then Todd must say the next two numbers (2 and 3), then Tucker must say the next three numbers (4, 5, 6), then Tadd must say the next four numbers (7, 8, 9, 10), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number 10,000 is reached. What is the 2019th number said by Tadd?
(A) 5743
(B) 5885
(C) 5979
(D) 6001
(E) 6011
Are You Up For the Challenge: 700 Level Questions
(A) 5743
(B) 5885
(C) 5979
(D) 6001
(E) 6011
Are You Up For the Challenge: 700 Level Questions
11 Jun 2022, 01:56
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Round 1 - Tadd - 1 Todd - 2-3 Tucker - 4-6
Round 2 - Tadd - 7-10 Todd - 11-15 Tucker - 16-21
Round 3 - Tadd - 22-28 Todd - 29-36 Tucker - 37-45
Round 4 - Tadd - 46-55 Todd - 56-66 Tucker - 67-78
We can see that, Tadd says 1 number in round 1, 4 numbers in round 2, 7 numbers in round 3, and in general 3n-2 numbers in round n. At the end of round n, the number of numbers Tadd has said so far is 1 + 4 + 7 + ....... + (3n-2) = \(\frac{{n(3n-1)}}{2}\), from the formula of sum of arithmetic series.
We find by trying value testing, that \(\frac{{37(110)}}{2}\)=2035, so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37 since in round 37, Tadd says (3*37)-2 or 109 numbers. At the end of Tadd's turn in round 37, the children have, in total, completed 36+36+37=109 turns. In general, at the end of turn n, the nth triangular number is said \(\frac{{n(n+1)}}{2}\). So at the end of turn 109 (that is the end of Tadd's turn in round 37), Tadd says the number \({109(110)}*2\)=5995. Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was 5995-16=5979.
Thus, the answer is (C) 5979.
Round 2 - Tadd - 7-10 Todd - 11-15 Tucker - 16-21
Round 3 - Tadd - 22-28 Todd - 29-36 Tucker - 37-45
Round 4 - Tadd - 46-55 Todd - 56-66 Tucker - 67-78
We can see that, Tadd says 1 number in round 1, 4 numbers in round 2, 7 numbers in round 3, and in general 3n-2 numbers in round n. At the end of round n, the number of numbers Tadd has said so far is 1 + 4 + 7 + ....... + (3n-2) = \(\frac{{n(3n-1)}}{2}\), from the formula of sum of arithmetic series.
We find by trying value testing, that \(\frac{{37(110)}}{2}\)=2035, so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37 since in round 37, Tadd says (3*37)-2 or 109 numbers. At the end of Tadd's turn in round 37, the children have, in total, completed 36+36+37=109 turns. In general, at the end of turn n, the nth triangular number is said \(\frac{{n(n+1)}}{2}\). So at the end of turn 109 (that is the end of Tadd's turn in round 37), Tadd says the number \({109(110)}*2\)=5995. Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was 5995-16=5979.
Thus, the answer is (C) 5979.
14 Jun 2022, 04:07
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We need to arrange the data in a tabular form where Tadd, Todd and Tucker are the Column heads and Row 1, 2, 3 etc are the rows where we need to enter the numbers belonging to Tadd/Todd/Tucker.
By doing so we'll se the distribution of numbers against Tadd as follows-
1. 1 nos. only (1)
2. 4 nos. (7 - 10)
3. 7 nos. (22 - 28)
4. 10 nos. (46 - 55)
... and so on.
The number of numbers in each row of Tadd follow a pattern, i.e., 1 nos, 4 nos, 7 nos, 10 nos,...[1+3(n-1)].
Now we have to identify the 2019th number said by Tadd. We can get close to it by adding the number in each rows.
Since it's following an AP,
S = (n/2)*[2+3(n-1)]
This sum should be close or equal to 2019.
We ll see that at n=35, S = 1820
n=36, S = 1926 (meaning, sum of the number of numbers till 36th row is 1926)
Number of numbers in 36th row is [1+3(36-1)] = 106.
Therefore we know that the 2019th number has to be in 37th row only.
Now lets track the pattern of last number of each row. These numbers are 1, 10, 28, 55...
Nth number is = 1 + 9 + 2*9 + 3*9...
= 1 + 9 [1+2+3+4 ... ... (n-1)]
= 1 + 9 [(n-1)(n)]/2 ... Where n is the row number.
To identify the 1926th number or the last number of 36th row we replace n by 36,
= 1 + 9(35)(36)/2
= 5671
Now if we add 107 we'll get the last number of Todd and further addition of 108 will give last number of Tucker (in 36th row only).
= 5671+107+108 = 5886
=> 1927th number by Tadd or the 1st number in 37th row is 5887.
Difference in 2019 and 1927 is = 92.
Therefore, 2019th number is = 5887 + 92 = 5979.
Answer (c)
Posted from my mobile device
By doing so we'll se the distribution of numbers against Tadd as follows-
1. 1 nos. only (1)
2. 4 nos. (7 - 10)
3. 7 nos. (22 - 28)
4. 10 nos. (46 - 55)
... and so on.
The number of numbers in each row of Tadd follow a pattern, i.e., 1 nos, 4 nos, 7 nos, 10 nos,...[1+3(n-1)].
Now we have to identify the 2019th number said by Tadd. We can get close to it by adding the number in each rows.
Since it's following an AP,
S = (n/2)*[2+3(n-1)]
This sum should be close or equal to 2019.
We ll see that at n=35, S = 1820
n=36, S = 1926 (meaning, sum of the number of numbers till 36th row is 1926)
Number of numbers in 36th row is [1+3(36-1)] = 106.
Therefore we know that the 2019th number has to be in 37th row only.
Now lets track the pattern of last number of each row. These numbers are 1, 10, 28, 55...
Nth number is = 1 + 9 + 2*9 + 3*9...
= 1 + 9 [1+2+3+4 ... ... (n-1)]
= 1 + 9 [(n-1)(n)]/2 ... Where n is the row number.
To identify the 1926th number or the last number of 36th row we replace n by 36,
= 1 + 9(35)(36)/2
= 5671
Now if we add 107 we'll get the last number of Todd and further addition of 108 will give last number of Tucker (in 36th row only).
= 5671+107+108 = 5886
=> 1927th number by Tadd or the 1st number in 37th row is 5887.
Difference in 2019 and 1927 is = 92.
Therefore, 2019th number is = 5887 + 92 = 5979.
Answer (c)
Posted from my mobile device
