Solution

Given:
• K = 1 * 2 + 2 * 3 + 3 * 4 + … + 21 * 22

To find:
• The remainder when K is divided by 23

Approach and Working:
If we observe the terms of the series K, we can generalise it as follows:

• 1st term = 1 * 2
• 2nd term = 2 * 3
• 3rd term = 3 * 4
• Hence, nth term = n * (n + 1) = \(n^2 + n\)

Now, sum of all the terms of K = \(∑ n^2 + ∑ n\) = \(\frac{{n (n + 1) (2n + 1)}}{6} + \frac{n (n + 1)}{2}\)
Simplifying, we get K = \(\frac{{n (n + 1) (n + 2)}}{3}\)

• If n = 21, K = \(\frac{21 * 22 * 23}{3}\)

o Therefore, it will be always divisible by 23, giving a 0 remainder

Hence, the correct answer is option E.

Answer: E
_________________

Register for free sessions
Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Must Read Articles
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Hey jenks88,
We didn't derive the formula for sum of all terms of k. We figured out the general expression representing any term of the series k (which is \(n^2 + n\)).

Once we have the general term, we applied the formula to find the sum of those elements forming the general term, to get the sum of all elements of k.

\(∑n^2 = \frac{n (n + 1) (2n + 1)}{6}\)

\(∑n = \frac{n (n + 1)}{2}\)

Hope this answers your query.
_________________

Register for free sessions
Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Must Read Articles
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Hey jenks88 ,

I would never prefer to bother about how this formula is derived. I can directly use this formula whenever required.

But if you are still curious to know how we arrived at this formula, you can refer here : http://www.math-only-math.com/sum-of-th ... mbers.html
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place
Blog: Subscribe to Question of the Day Blog
GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.

Hi Jenks88,

I would suggest to memorize the below formulae :-

Arithmetic progression:-

1. 1 + 2 + 3 + ...... + n = [n(n + 1)]/2.

2.1² + 2² + 3² +……………. + n² = [n(n+ 1)(2n+ 1)]/6.

3.1³ + 2³ + 3³ + . . . . + n³ = [{n(n + 1)}/2 ]

Geometric Progression:-

(i) The general form of a G.P. is a, ar, ar², ar³, . . . . . where a is the first term and r, the common ratio of the G.P.

(ii) The n th term of the above G.P. is t₀ = a.rn−1 .

(iii) The sum of first n terms of the above G.P. is S = a ∙ [(1 - rⁿ)/(1 – r)] when -1 < r < 1

or, S = a ∙ [(rⁿ – 1)/(r – 1) ]when r > 1 or r < -1.

(iv) The geometric mean of two positive numbers a and b is √(ab) or, -√(ab).

(v) a + ar + ar² + ……………. ∞ = a/(1 – r) where (-1 < r < 1).

Also, practice several sums on summation & series so that you would be able to master the application of the above formulae.

Hope it helps.
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Hey GyMrAT,
I have a query regarding the highlighted portion.
If we take a similar example, let's say
17 - 12 is a multiple of 5, does that necessarily mean both 17 and 12 are divisible by 5?
_________________

Register for free sessions
Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Must Read Articles
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Beside solution of EgmatQuantExpert but you have to memorize the formula of the sum of consecutive squares, I want to provide the solution without memorizing that formula:

\(S = 1 \times 2 + 2 \times 3 + {...} + (n-1) \times n\)

Note that \(3 \times k \times (k+1) = [(k+2) - (k-1)] \times k \times (k+1) = -(k-1)k(k+1) + k(k+1)(k+2)\)

So we have

\(3S = 3 \times 1 \times 2 + 3 \times 2 \times 3 + {...} + 3 \times (n-1) \times n\)
\( = - (0 \times 1 \times 2 )+ (1 \times 2 \times 3) - (1 \times 2 \times 3) + (2 \times 3 \times 4) - {...} - [(n-2)(n-1)n] + [(n-1)n(n+1)]\)
\(= (n-1)n(n+1)\)

Hence \(S= \frac{(n-1)n(n+1)}{3}\)

Thus \(K = \frac{21 \times 22 \times 23}{3} = 7 \times 22 \times 23\). Answer E
_________________

Actual LSAT CR bank by Broall

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

New Error Log with Timer