If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder

Question

If  22^3+23^3+24^3+....+87^3+88^3  is divided by 110 then the remainder will be

A. 55
B. 1
C. 33
D. 0
E. 44

jax91 · Answer

(a^n + b^n ) is divisible by a+b , given that n is odd.

we need to form pairs of numbers whose sum is 110.

(22^3 + 88^3 ) + (23^3 + 87^3 ) ...... + 55^3

so all pairs are divisibly by 110 , which leave 55^3/110 , and the remainder for this is 55.

so IMO A.

jade3 · Answer

Correct Jax91. the answer is 55

kalpeshchopada7 · Answer

I wud differ with the final answer. if jade3's solution method is agreed upon, we get 55^3/110. which is (55*55*55)/(55*2) or (55*55)/2. Thus the odd number divided by 2 will result into remainder 1.

In that case the OA should B.

jade3 · Answer

The colored step cannot be used because, while finding the remainder you cannot reduce the numbers(Ex:when you divide 32/64 the remainder is 32, But if you reduce 32/64 to 1/2 you get the remainder 1)

According to remainder theory

If P be the product of N1, N2, N3…

And let D divide P, then the remainder of P/D would be same as remainder of (Mod *Mod *Mod *…..)/D where Mod denotes the remainder operation

In our case Remainder of (55*55*55/110)= Remainder of (3025*55/110)
Remainder of(3025/110) = 55
Remainder of(55/110)=55
Now according to the remainder theorem Remainder of (55*55*55/110)= Remainder of (55*55/110)=55

kalpeshchopada7 · Answer

Thanks Jade3. I acknowledge my mistake. +1 from me!!

ezinis · Answer

Can somebody prove this statement for me please:
(a^n + b^n ) is divisible by a+b , given that n is odd.

Thanks

kalpeshchopada7 · Answer

a^3+b^3 = (a+b)(a^2-ab+b^2).
a^5+b^5 = (a+b){a^4 - (a^3)b + (a^2)(b^2) - a(b^3) + b^4)}
....
Like wise, 
a^n+b^n   = (a+b){a^(n-1) -  *b +  *b^2 +.....+ b^n-1}

Ezinis, Guess this should be clear. You can try solving for yourself.

GMAT TIGER · Answer

Given that n is odd:

(a^n + b^n ) = a^(n-1) - a^(n-2)b + a^(n-3)b^2 - a^(n-4) b^3 + ........... - ab^(n-1) + b^n

Suppose n = 5: 
(a^5 + b^5 ) = (a + b) (a^4 - a^3b + a^2b^2 - ab^3 + b^4)

Suppose n = 7: 
(a^7 + b^7 ) = (a + b) (a^6 - a^5b + a^4b^2 - a^3b^3 + a^2b^4 - ab^5 + b^6)

jade3 · Answer

Let us try backwards
Let n be odd then

(x+y)(x^(n-1) - x^(n-2) y + ... --x y^(n-2) + y^(n-1)) = (x^n-x^(n-1)y+……..-x y^(n-1)+y^n)=x^n+y^n
Because, you will notice that most of the terms would cancel in pairs; and the result would be x^n+y^n

You can try for n=3

(x+y)(x^2-x y+y^2)=x^3-x^2 y+ x y^2+ y x^2-y^2 x+y^3

You notice that the second term and fourth term cancel each other 
Similarly You notice that the third term and fifth term cancel each other

You get the result as x^3+y^3

Now can try for x=5 yourself.

ezinis · Answer

thanks guy +1. good one to remember

amit2k9 · Answer

interesting one indeed.

jantwarog · Answer

I don't get how you all guys get the remainder of 55^3/110 without any complicated calculations. Can someone explain?

If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards