If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai

Question

If  x = (16^3+17^3+18^3+19^3) , then x when divided by 70 leaves a remainder of:

A. 0 
B. 1 
C. 69 
D. 35 
E. 42

jaysonbeatty12 · Accepted Answer

This question can be solved by focusing on only the units digit of x and the answer choices. The units digit of the answer choices are all unique. So, by knowing only the units digit of x we can know the units digit of the right answer.

This would NOT work if the units digit of the answer choices were all the same - in this case say 0, 10, 20, 30, and 40. But, of course, the GMAT is a test of your ability to think and simplify, not of your ability to do a bunch of arithmetic, so this is the type of thing you want to be on the lookout for.

So in this case

The units digit of 16 cubed is 6
The units digit of 17 cubed is 3
The units digit of 18 cubed is 2
The units digit of 19 cubed is 9

6 + 3 + 2 + 9 = 20. So, x has a units digit of 0. So, the remainder of x divided by 70 will also have a units digit of 0. A is the only answer choices that works.

A is correct.

Think before you math

KarishmaB · Answer

We know:
 a^3 + b^3 = (a + b)*(a^2 - ab + b^2)

16^3 + 19^3 = (35)(16^2 - 16*19 + 19^2) 
 17^3 + 18^3 = (35)(17^2 - 17*18 + 18^2)

Therefore,  (16^3+17^3+18^3+19^3) = 35(16^2 - 16*19 + 19^2 + 17^2 - 17*18 + 18^2)

(16^2 - 16*19 + 19^2 + 17^2 - 17*18 + 18^2)  is a multiple of 2 since we have (Even - Even + Odd + Odd - Even + Even). Hence the entire expression is Even.
Therefore, x is divisible by 70.

Why did I think of algebraic identities? - because exponents were 3 for all the terms and there seemed to be no +1, -1 symmetry. The numbers weren't close to a multiple of 70 and were too big. Most people don't know the squares/cubes of 16, 17 etc.

sudhir18n · Answer

Hi Ankit, I dont think GMAT will ask this , but to answer your question , the best way to do this is
a^+b^3 = (a+b)(a^2 -ab +b^2)
( 16^3+19^3) + (17^3+18^3)
(16+17)( 16^2 -16*19+19^2) + (17+18)(17^2-17*18 +18^2)

35*( 16^2 -16*19+19^2) (17^2-17*18 +18^2)

.
so we can re write the equation as 
35^(even -even+odd)(even-even+odd)
odd*odd = even = atleast one 2

35*2 (...........) (...........)
 when this equation is divided by 70 , we will have 0 as the remainder.

hope this helps.

yogesh1984 · Answer

A good to know question IMO but still I would take it this way- At first glance I read a symmetry here as I break the A^3 + B^3 formula- pair of 16, 19 and of 17,18 give me a total of 35 (something useful considering I am looking for divisibility by 70. Now my job is half done but I have still to check whether the remaining is even - to check for divisibility by 2). after opening the formula A^3 + B^3 nd rearranging terms one would get- Lets recall - (A^3 + B^3) = (A+B)(A^2 + B^2 - A*B) .. so pair 1 gives- (16^2 + 19^2 - 16*19 ) (16+19) = 35 * (16^2 + 19^2 - 16*19 ) pair 2 gives- (17^2 + 18^2 - 17*18) (17+18) = 35 * (17^2 + 18^2 - 17*18) now lets add them, so we have got now- (35) * (16^2 + 17^2 + 18^2 + 19^2 - 16*19 - 17*18) - to get down finally we need to determine whether (16^2 + 17^2 + 18^2 + 19^2 - 16*19 - 17*18) is divisible by 2 - (Even + Odd + Even + Odd - Even - Even). You see we are done here (obviously remainder is big ZERO :!:

PKN · Answer

Note:-  a^n+b^n  is divisible by (a+b) when n is ODD.

In line with the above,  a^3+b^3  is divisible by (a+b)

a)  17^3 + 18^3  is divisible by 17+18=35
b)  16^3 + 19^3  is divisible by 16+19=35
c) Sum of unit digits of 17^3 + 18^3, 3+2=5, Sum of unit digits of 16^3 + 19^3, 6+9=15. So the given expression is an EVEN number

Hence from (a),(b), and (c), we have 
 x = (16^3 + 17^3 + 18^3 + 19^3) = 17^3 + 18^3+16^3 + 19^3=35k+35p=35*2*y=70y 
So, the given expression when divided by 70 leaves a remainder 0.

Ans. (A)

Another approach:-
 a^n+b^n+c^n+d^n  is divisible by (a+b+c+d) when n is ODD , a, b, c, and d are in A.P.(Arithmetic progression)
Here 16,17,18, and 19 are in AP. n=3, which is ODD’
Hence,  x = (16^3 + 17^3 + 18^3 + 19^3)  is divisible by (16+17+18+19=70)

Therefore, remainder is zero.

abhinav5 · Answer

a^3+b^3=(a+b)(a^2+b^2-ab)
16^3+19^3=35k
17^3+18^3=35k
So x=35k +35k=70k
1) is the answer

Mansoor50 · Answer

Hi....

would the following logic work?

16^3 + 17^3 is iv by 33
18^3 + 19^3 is div by 37

therefore

16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70

regards

KarishmaB · Answer

Think about it:

a is divisible by 3 (say a = 6)
and b is divisible by 5 (say b = 5)

Does this mean (a + b = 11) is divisible by 8? It is certainly not necessary.

Mansoor50 · Answer

Thanks!!! i should have tested the assumption myself....!!

Fdambro294 · Answer

A bit of a different way of analyzing the problem.

We can break up the divisor into its prime factors and look for a number that has the same remainder properties as X = 16^3 + 17^3 + 18^3 + 19^3 when it is divided by 70

Divisor 70 = 2 * 5 * 7

(1st) what remainder does X yield when divided by 7?

(16’3 /7)Rof + (17’3 /7)Rof + (18’3 /7)Rof + (19’3 /7)Rof = excess remainder when X is divided by 7

= (2)’3 + (3)’3 + (-3)’3 + (-2)’3

= 8 + 27 - 27 - 8 = 0

X is a number that yields a remainder of 0 when divided by 7.

(2nd)divide X by 5. We can use the Units Digit of each portion to find the remainder and Add the remainders

16’3 = .....6 ——— Rem of 1

+

17’3 = .....3 ——— Rem of 3

+

18’3 = ....2 ——— Rem of 2

+

19’3 =......9 ——- Rem of 4

1 + 3 + 2 + 4 = excess remainder of 10

10/5 —— Rem of 0

X is a number when divided by 5 yields a remainder of 0

(3rd) divide X by 2

Using the Even/odd nature of the values and the fact that an exponent does not change the even/odd nature of a given number:

X = (even) + (odd) + (even) + (odd) = EVEN integer result

X is a number that leaves a remainder of 0 when divided by 2

Thus: X = 7a = 5b = 2c

Which means X is:

-multiple of 7

-multiple of 5

And multiple of 2

By the factor foundation rule, this means that X is also a multiple of 70.

The remainder is 0

Answer (A)

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Iq · Answer

I am not sure if this method is correct, but this is how I did it:

x =  16^3 + (16 + 1)^3 + (16 + 2)^3 + (16+3)^3

for 70 => 7 x 10, so first tried to find remainder when dividing by 7,

for  16^3  mod 7,  remainder is 1

also 16 mod 7 = 2

so,  (16 + 1)^3  mod 7 =  3^3  mod 7 =  6

Similarly, for the  (16 + 2)^3, (16 + 3)^3 , remainders when divided by 7 are  1, 6  resp.

Adding all remainders: 1 + 6+ 1 + 6 = 14 and 14 mod 7 = 0

and now, dividing by 10 also, the remainder stays 0.

Ans  A

VeritasKarishma  could you please comment if this method is correct

KarishmaB · Answer

Yes, since 7 and 10 are co-prime, if you get the same remainder on division by both, that will be the remainder on division by 70 too.

AakankshaArora · Answer

If we are checking remainder for 70, shouldn't we check for both 7 and 10. 20 is divisible by 10 but has a different remainder with 7

Tomsapo · Answer

Wouldn't this way to solve mean that any number ending with 0 as x would get a 0 remainder when divided by 70? This would mean that for example if we have x=1210 it should be divisible by 70, but this is not the case.

Krunaal · Answer

If x = 1210; 1210/70 will give you remainder 20 - the last digit here is  0 . That's the reason they mention that the method would NOT work if the units digit of answer choices were all same - 0, 10, 20, 30, and 40.

When x is divided by any multiple of 10, the  units digit of the remainder   is same as the units digit of x.

bumpbot · Answer

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If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai

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