The answer is 10.

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We first find the row. The sum of the number of values of the first $n$ rows is the sum of the integers representing those rows; thus the total number of values in rows $n$ and above is $\binom{n+1}{2}$ . We are then looking for the closest row to the $1000^{th}$ number, so we approximate this by $\binom{n+1}{2}\approx1000\rightarrow n^2-n\approx 2000 \rightarrow n^2-n-2000\approx0$

We then use the quadratic formula and take the positive solution to get $n \approx \frac{1+3\sqrt{889}}{2}\approx 44$ (note that even though we have taken $\lfloor \sqrt{889} \rfloor$ along with our original approximation of $\binom{n+1}{2}\approx 1000$ , the second approximation does not affect the first and vice versa, meaning our approximation is still accurate.) This implies the last complete row before the $1000^{th}$ number is $44+1=45$ . We have $\binom{45}{2}=990$ , so the last value of row $45$ is also the $990^{th}$ number.

Thus the $1000^{th}$ number is the $10^{th}$ number of row $46$ , or $\boxed{10}$

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in row 1, we have 1 integer (1)

in row 2, we have 2 integers (1,2)

in row 3, we have 3 integers (1,2,3)

......

in row 1000, we have 1000 integers (1,2,3,...,1000)

So, take the number of integers in row n to form an arithmetic progression (AP)

AP: 1,2,3,...,1000

To determine the row of 1000th number,

S <= 1000

(n/2) [2a+(n-1)d] <=1000

(n/2) [2x1+(n-1)1] <=1000

(n/2) (2+n-1) <=1000

(n/2) (n+1) <=1000

n^2+n<=2000

n^2+n-2000<=0

n=44.23 or -45.22

n denotes the number of row, so n cannot b negative, therefore n=44.23

This n value represents 1000th number is in the 45th row

To determine where the 1000th number in the 45th row, substitute n=44 into

(n/2) (n+1)=990

1000-990=10

The 1000th number is the 10th element in the 45th row, therefore the 1000th number is 10

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this is a "Pattern Recognition" group sequence, let transform the problem to: what is the value at 1000th term, of the sequence: (1), (1,2), (1,2,3), ...

First: we find the row contains the 1000th term by using n(n+1)/2 = 1000 n = 44, so the 1000th term in group 44 + 1 = 45

Second: max value of group 44 is 44*45/2 = 990 then in group 45, the value of 1000th is 10 (because 1000 - 990 = 10)

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numbers of integer at row $i$ = $i$

At row 1, we got 1 integer(s); at row 2, we got 2; at row 3 we got 3....

This form an A.P. with $a=1$ and $d=1$

For the 1000th integer, we must find the number of row containing it.

From sum of A.P formula, $\frac{n}{2} (1 + n) = 1000$ ,

solving it and we get $n = 44.22$ or $n = -45.22$ ,

reject the negative answer since in n is always positive.

Hence, we get $n = 44.22$ , meaning that the 1000th integer is at row 45.

Up to row 44, the number of integers is $1 + 2 + 3 + ... + 44 = 990$ and it's 10 more to go to the 1000th integer.

Hence, the 1000th integer(or 10th in row 45) is $10$

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You should place the $n+1$ term in parentheses as one would expect $\frac{n + 1 \times n}{2}$ to evaluate to $\frac{n + n}{2} = n$ .

Drew Cummins
- 7 years, 9 months ago

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Agreed! Order of operations help us avoid ambiguity in our statements.

Sorry about the mistake.

Tan Li Xuan
- 7 years, 9 months ago

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Pl see

A Joshi
- 7 years, 9 months ago

Use Gauss law to find that sum of integers from 1 to 45 is 990. Ans 10

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Realise that the total number of numbers read out after N rows is the sum of the series

$1+2+3....+n$ , i.e the sum of the triangle numbers, which is an arithmetic series, which has the summation formula:

$S_{N}=\frac{N(N+1)}{2}$

So we can work out what row the 1000th number is in by finding the N such that $S_{N}=1000$

Meaning:

$N^{2}+N-2000=0, N=\frac{-1±\sqrt{8001}}{2}$

$90^{2} =8100 => \sqrt{8001} \sim 89 => N \sim 44$

So 1000 is part of the way into the 45th row of the triangle. There are 990 numbers up to the 44th row, implying that the 1000th number read is 10 into the 45th row, so is 10.