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

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.

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

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.

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

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

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.

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

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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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

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.

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

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