M4maths Previous Puzzles

01 Jun 2016 Remainder

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

OPtion

1) 0
2) 1
3) 69
4) 3
5) 54
6) 8
7) 5
8) 43
9) 17
10)62

Solution

X = ( 16^3 + 17^3 + 18^3 + 19^ 3 )
(a+b) is a factor of (a^3+b^3)
(16^3+19^3) is divisible by 16+19 = 35
(17^3+18^3) is divisible by 17+18 = 35
so,X = ( 16^3 + 17^3 + 18^3 + 19^ 3 ) has a factor 35
also, X = ( 16^3 + 17^3 + 18^3 + 19^ 3 )is an even no. (unit digit is 6+3+2+9= 0) so, X has a factor 2.
[since, E+O+E+O = E]
so, X is completely divisible by 35*2 = 70 & leaves a remainder 0.
Option 1)

Correct Option: 1 Best Solution (16)

Mukund Manohar   8 years AGO

16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)
= 35*(odd number) + 35*(odd number)
= 35*(even number)
=70 K,where K is a natural number
when 70K divided by 70,remainder = 0

Like? Yes (6) | No   

Lakshmipathi   8 years AGO

very simple sum of base values divide by the number 16+17+18+19=70
70%70=0

Like? Yes (6) | No   

Kawarpreet Singh   8 years AGO

As 16+17+18+19=70, then the sum of their cubes will be perfectly divided by 70.....
Hence remainder will be zero (Choice 1)

Like? Yes (6) | No (1)  

ABHINAV PUSHPAM   8 years AGO

16 + 17 + 18 + 19 = 70

x = 70k (k is any positive number)

Thus, x is divisible by 70.

So, remainder will be 0

Option 1)

Like? Yes (9) | No (5)  

shreyas kj   8 years AGO

Instead of going for obvious formula, (a + b) % c = a%c + b%c, try to calculate the value of above series by formula :-
1^3 + 2^3 + .... + n^3 = (n * (n+1) / 2) ^ 2
Since the above series starts from 16^3, above formula cannot be applied directly.
Looking at above series, it can be seen as :-
(1^3 + 2^3 + ... + 18^3 + 19^3) - (1^3 + 2^3 + ..... + 14^3 + 15^3)
Now we have two series, applying formula in each series, we have:-
(19 * 20 / 2) ^ 2 - (15 * 16 / 2) ^ 2
=> 190 ^ 2 - 120 ^ 2
=> (190 + 120) * ( 190 - 120) -----> a^2 - b ^2 = (a+b)*(a-b)
=> 310 * 70
Series is divisible by 70 hence option 1
Note :- I tried it on scientific calculator and it gave answer without decimal point. You can try it too.

Like? Yes (1) | No   

Rajeet yadav   8 years AGO

by using formula
(a+b)^3 =a3 +b3+ 3ab(a+b)
consider 16 and 19as one and 17& 18 as lther

Like? Yes | No   

lucky chaudhary   8 years AGO

16^3=4096 ,17^3=4913 ,18^3=5832, 19^3=6859 4096+4913+5832+6859=21700 WHEN 21700 IS DIVIDED BY 70 ,THE REMINDER IS 0

Like? Yes (1) | No (1)  

Akshat Sood   8 years AGO

(16)�=(70-54)�
(17)�=(70-53)�
...
..
...So on till 19

on dividing these no by 70 we eliminate 70 from above numbers. we are left with 54,53,52,51

Sum if these are divisible by 70 hence remainder is 0

Like? Yes | No   

Raghav   8 years AGO

16 ^ 3 = 4096
17 ^ 3 = 4913
18 ^ 3 = 5832
19^ 3 = 6859

all adding and divide by 70 the remainder is 0

Like? Yes | No   

PRASAD CHAVAKULA   8 years AGO

the cubes of 16 and 17 and 18 and 19
total 21700/70
so reminder 0

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rahul Tiwari   8 years AGO

17^3+19^3) is a % by 2 so sum is di by 35 so totally its % by 70 so zero rem

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Ankita   8 years AGO

Since it completely divisible, it gives remainder as 0
Thus, option 1

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Rega   8 years AGO

16*16*16+17*17*17+18*18*18+19*19*19/70 =21700/70 = 310 . Reminder is 0

Like? Yes | No   

Adhita kundu   8 years AGO

16+18+17+19=70
X=70k(k is any +ve integer)
Thus x is divisible by 70

Like? Yes | No   

viswambher   8 years AGO

X = 21700
X/70 = 310

remainder is 0

Like? Yes | No (2)  

kallam ravi sankar readdy   8 years AGO

21700 divides with 70 we get 310 as quotient and remainder 0

Like? Yes (1) | No (3)