In an infinite sequence, a_n is the remainder when n^2

Question

In an infinite sequence,  a_n  is the remainder when  n^2  is divided by 10. Which of the following cannot be the sum of the first  k  terms of the sequence, where  k  is a positive integer?

(A) 1,115
(B) 1,126
(C) 1,170
(D) 1,195
(E) 1,291

GridLock · Answer

Start by creating the sequence

n = 1 2 3 4 5 6 7 8 9 10
n^2 = 1 4 9 16 25 36 49 64 81 100
n^2/10= 1 4 9 6 5 6 9 4 1 0

By now, I realized that focusing on units digit can help with elimination

Unit digit of sum of first k terms = 1 5 4 0 5 1 0 4 5 5
Basically only these 4 digits are repeating 1, 5, 4, 0

Therefore eliminate all except (A) which ends in 6

kevincan · Answer

1,5,4,0,5,1,0,4,5,5,6,...

There is a 6 at the 11th cumulative sum.

kevincan · Answer

I have modified the answer choices

GridLock · Answer

Though I feel, this method is very lengthy, this is the only approach I could think of. Not sure if I would attempt this in the actual exam.

n = 1 2 3 4 5 6 7 8 9 10
n^2 = 1 4 9 16 25 36 49 64 81 100
n^2/10= 1 4 9 6 5 6 9 4 1 0
Sum = 1 5 14 20 25 31 40 44 45

This pattern shall repeat for every 10 terms.

The sum of remainders can be expressed by
S = 45*n +

i.e when Sum is divided by 45, the remainder must be one of the above.

1115/45 R = 35 => Not valid
1126/45 R = 1 => Valid
1170/45 R = 0 => Valid
1195/45 R = 25 => Valid
1291/45 R = 31 => Valid

Therefore, answer is option A

kevincan · Answer

What part did you find the most time consuming ?

GridLock · Answer

Reading/understanding question ~40 seconds

Writing the series of n^2, n^10, its sums itself is going to take atleast 90s

So far I wouldn't even know I am going to get an answer;

In exam, under time pressure, I might have bookmarked and skipped the question.

Checking individually for each answer type is another time consuming point; To divide each answer choice by 45 and finding the remainder is going to be tremendous task specially if the correct answer choice was placed at (E) rather than (A)

kevincan · Answer

Luckily , the answer choices are closely spaced : suppose you started with C

1170/45 =26 remainder 0
1126 = 1170 -44 , so this yields a remainder of 1
1115= 1126-11, so this yields a remainder of 35

GridLock · Answer

Makes sense, thanks for the tip!

kevincan · Answer

When you’ve done a lot of these questions, you realize that you should always look at the answer choices for clues or shortcuts. Great explanation otherwise by the way !

HarshavardhanR · Answer

n 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
 10 
 11 
 12 
 13 
 
   n^2  
 1 
 4 
 9 
 16 
 25 
 36 
 49 
 64 
 81 
 100 
 121 
 144 
 169 
 
  An 
 1 
 4 
 9 
 6 
 5 
 6 
 9 
 4 
 1 
 0 
 1 
 4 
 9

Observe:

The pattern for An,  { 1-4-9-6-5-6-9-4-1-0 }  repeats every 10 terms.

Sum of 1st ten terms = 1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0 = 45.

So, as far as the infinite sequence is concerned:

- Sum of 1st 10 terms = 45
- Sum of 1st 20 terms = 2 x 45 = 90

Now, observe the choices. How do we get close to these numbers using blocks of ten numbers (summing to 45)?

- 45 x 20 blocks of ten terms= 900
- 45 x 30 blocks of ten terms = 1350

So, let's try 25 blocks of ten terms

- 45 x 25 = 1125. Super close to choice (B).

What the above means:

Sum of the sequence till 250 terms (25 blocks x 10 terms/block) = 1125

We can quickly check choices (A) and (B) from here.

Term 
 246th 
 247th 
 248th 
 249th 
 250th 
 251st 
 
  Sum till that term 
  1111  
  1120  
 1124 
 1125 
  1125  
  1126

Observe:

- Choice B? Possible
- Choice A? Not possible. From 1111, the sum of the sequence moves to 1120 (no 1115).

There can only be one right answer. Choice A is the correct answer!

---
Harsha

kevincan · Answer

Thanks for taking the time to write this fantastic explanation !

In an infinite sequence, a_n is the remainder when n^2

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n	1	2	3	4	5	6	7	8	9	10	11	12	13
\(n^2\)	1	4	9	16	25	36	49	64	81	100	121	144	169
An	1	4	9	6	5	6	9	4	1	0	1	4	9

Term	246th	247th	248th	249th	250th	251st
Sum till that term	1111	1120	1124	1125	1125	1126

GMAT Problem Solving (PS) Questions

In an infinite sequence, a_n is the remainder when n^2

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