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Algebra Level 3

Find the smallest positive integer a a such that a + ( a + 1 ) + ( a + 2 ) + … = 2021 a+(a+1)+(a+2)+ \ldots = 2021 .

To clarify, there is finitely many terms on the left-hand side of the equation.


The answer is 20.

Ossama Ismail by Ossama Ismail , 63, Egypt

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2 solutions

Ossama Ismail
Dec 20, 2020

a + ( a + 1 ) + ( a + 2 ) + ⋯ + ( a + n ) = ( 2 a + n ) ( n + 1 ) 2 = 2021 {a+(a+1)+(a+2)+ \cdots+(a+n)} = \dfrac{(2a+n)(n+1)}{2} =2021

( 2 a + n ) ( n + 1 ) = 4042 = 2 ∗ 43 ∗ 47 {(2a+n)(n+1)} =4042 = 2 * 43 *47

so, either ( 2 a + n ) = 86 and ( n + 1 ) = 47 ⟹ a = 20 (2a+n)=86 \text{ and } (n+1) = 47 \implies a = 20

or

( 2 a + n ) = 94 and ( n + 1 ) = 43 ⟹ a = 26 (2a+n)=94 \text{ and } (n+1) = 43 \implies a = 26

answer: a = 20 a=20

All other cases will give a > 20 or a < 0 a > 20 \text{ or } a < 0 .

You may check David Verkan's complete solution.

0 Helpful 0 Interesting 5 Brilliant 0 Confused

It should be made clear that you are asking for the smallest positive integer since the smallest positive number for a a is 0.078125 (0.078125 + 1.078125 + ... + 63.078125 = 2021).

Alexander McDowell - 5 months, 3 weeks ago

Ok. I will modify the question.

My question to you How did you find this value? thanks.

Ossama Ismail - 5 months, 3 weeks ago

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Your equation ( 2 a + n ) ( n + 1 ) 2 = 4042 \frac{(2a + n)(n + 1)}{2} = 4042 solves to a = 2021 n + 1 − n 2 > 0 a = \frac{2021}{n + 1} - \frac{n}{2} > 0 , so if a a is not limited to an integer but n n is, then 1 ≤ n ≤ 63 1 \leq n \leq 63 , so that smallest positive non-integer value of a a is a = 2021 63 + 1 − 63 2 = 5 64 a = \frac{2021}{63 + 1} - \frac{63}{2} = \frac{5}{64} .

David Vreken - 5 months, 3 weeks ago

For a complete solution, wouldn't you also need to consider ( 2 a + n ) = 2021 (2a + n) = 2021 and ( n + 1 ) = 2 (n + 1) = 2 , ( 2 a + n ) = 4042 (2a + n) = 4042 and ( n + 1 ) = 1 (n + 1) = 1 , ( 2 a + n ) = 2 (2a + n) = 2 and ( n + 1 ) = 2021 (n + 1) = 2021 , ( 2 a + n ) = 43 (2a + n) = 43 and ( n + 1 ) = 94 (n + 1) = 94 , ( 2 a + n ) = 47 (2a + n) = 47 and ( n + 1 ) = 86 (n + 1) = 86 , and ( 2 a + n ) = 1 (2a + n) = 1 and ( n + 1 ) = 4042 (n + 1) = 4042 ?

David Vreken - 5 months, 3 weeks ago
David Vreken
Dec 21, 2020

Rearranging the given equation:

a + ( a + 1 ) + ( a + 2 ) + . . . + ( a + n ) = 2021 a + (a + 1) + (a + 2) + ... + (a + n) = 2021

( n + 1 ) a + ( 1 + 2 + . . . + n ) = 2021 (n + 1)a + (1 + 2 + ... + n) = 2021

( n + 1 ) a + 1 2 n ( n + 1 ) = 2021 (n + 1)a + \frac{1}{2}n(n + 1) = 2021

2 a = 4042 n + 1 − n 2a = \frac{4042}{n + 1} - n

If a a is an integer, then n + 1 n + 1 must divide into 4042 = 2 ⋅ 43 ⋅ 47 4042 = 2 \cdot 43 \cdot 47 . There are 8 8 factors of 4042 4042 to consider:

n + 1 n + 1 n n a a
1 1 0 0 1 2 ( 4042 1 − 0 ) = 2021 \frac{1}{2}(\frac{4042}{1} - 0) = 2021
2 2 1 1 1 2 ( 4042 2 − 1 ) = 1010 \frac{1}{2}(\frac{4042}{2} - 1) = 1010
43 43 42 42 1 2 ( 4042 43 − 42 ) = 26 \frac{1}{2}(\frac{4042}{43} - 42) = 26
47 47 46 46 1 2 ( 4042 47 − 46 ) = 20 \frac{1}{2}(\frac{4042}{47} - 46) = 20
86 86 85 85 1 2 ( 4042 86 − 85 ) = − 19 \frac{1}{2}(\frac{4042}{86} - 85) = -19
94 94 93 93 1 2 ( 4042 94 − 92 ) = − 25 \frac{1}{2}(\frac{4042}{94} - 92) = -25
2021 2021 2020 2020 1 2 ( 4042 2021 − 2020 ) = − 1009 \frac{1}{2}(\frac{4042}{2021} - 2020) = -1009
4042 4042 4041 4041 1 2 ( 4042 4042 − 4041 ) = − 2020 \frac{1}{2}(\frac{4042}{4042} - 4041) = -2020

The smallest positive integer of a a is therefore a = 20 a = \boxed{20} .

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