What is the remainder when K is divided by 23?

Question

K = 1 * 2 + 2 * 3 + ... + 21 * 22

A. 22
B. 11
C. 6
D. 2
E. 0

EgmatQuantExpert · Accepted Answer

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

jenks88 · Answer

How do you derive the formula for sum of all terms of k?

EgmatQuantExpert · Answer

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.

jenks88 · Answer

I get the general term equation but still unsure where ∑ n^{2} = \frac{n(n+1)(2n+1)}{6}  comes from?

abhimahna · Answer

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 : https://www.math-only-math.com/sum-of-th ... mbers.html

EgmatQuantExpert · Answer

Thanks to abhimahna . Please make a note of the most used 3 cases as following: • Sum of first n natural numbers = ∑n = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2} • Sum of first n square numbers = ∑n^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} • Sum of first n cube numbers = ∑n^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = ^2 Also note that, you can use them only when the numbers are consecutive and start from 1 Hope it helps

PKN · Answer

Hi Jenks88,

I would suggest to memorize the below formulae :-

Arithmetic progression:-

1. 1 + 2 + 3 + ...... + n =  /2.

2.1² + 2² + 3² +……………. + n² =  /6.

3.1³ + 2³ + 3³ + . . . . + n³ =

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 ∙   when -1 < r < 1

or, S = a ∙  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.

GyMrAT · Answer

EgmatQuantExpert

Can we approach the question in the following manner:

Given K = 1*2 + 2*3 + 3*4+.......+ 21*22

Consider P = 1*2 + 2*3 + 3*4+......+ 21*22 + 22*23

P - K = 22*23 = a Multiple of 23

Hence P & K are both divisible by 23. Therefore when K is divided by 23, the remainder is 0.

Thanks,
GyM

EgmatQuantExpert · Answer

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?

GyMrAT · Answer

*FacePalm*, damn i am smoking some weird ****, i guess.

Thanks  EgmatQuantExpert , for kicking my ass.

broall · Answer

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 $3 \times k \times (k+1) =   \times k \times (k+1) = -(k-1)k(k+1) + k(k+1)(k+2)$

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) - {...} -   +  $
$= (n-1)n(n+1)$

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

broall · Answer

Now, go back to the formula of the sum of square consecutive. How we could find that formula? Here is it.

M = 1^2 + 2^2 + {...} + n^2

Note that $k^2 = k \times k = k \times (k-1 + 1) = k(k-1) + k$, we have

$M = 1^2 + 2^2 + {...} + n^2 = (0 \times 1 + 1) + (1 \times 2 + 2) + {...} + (n(n-1) + n)$
$= (1 \times 2 + 2 \times 3 + {...} + (n-1) \times n) + (1 + 2+ {...} + n)$
$= M1 + M2$

We could easily have $M2 = 1 + 2+ {...} + n = \frac{n(n+1)}{2}$

Now, we have to calculate M1.

From my previous post, I have 
$M1 = 1 \times 2 + 2 \times 3 + {...} + (n-1) \times n = \frac{(n-1)n(n+1)}{3}$

Thus, we have

$M = M1 + M2 = \frac{(n-1)n(n+1)}{3} +\frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6}$

bumpbot · Answer

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What is the remainder when K is divided by 23?

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