If x=\left(16^{3}+17^{3}+18^{3}+19^{3} \right), then x divided by 70 leaves a remainder of what?
62first observe that this number is even
Then look at 163+193 this is divisible by 16+19 = 35
similarly for 173+183
Hence the number is divisible by 2 and by 35
hence by 70
Hence the remainder is zero.
1one other way which is the long way .
x= \sum_{1}^{19}{n^{3}} - \sum_{1}^{15}{n^{3}}
= 21700
which is divisible by 70 .
but Nishant Sir is better .
this question was asked in Thursday's Class 11th NSO.
1Arrey Yaar wat i've done is ..... just find out the last digit of the expression of the sum of the cubes...it'll come out to be 0 . and so it has to leave a remainder 0 when divided by 70...and there is 0 as an option. If u don't wanna try this mcq method ..go for this :
(13+23+33+......+193) - (13+23+......+153) Niow put the general formula : (n(n+1)2)2. I'm sure you will get somethingx70 as the answer....which is divisible by 70 .
11Gr8 job, aveek......
But a small confusion, 540 also ends with 0, but 70 unfortunately does not divide it! [2]
1Oyeeee Soumik...u hav options na..in MCQ ???? once u get last digit 0 and u've 0 as one and only one option...that's the most possible one naa ???? U've 4 options....0,1,2and 3....which one d'ya think is the most suitable ???
11Don't hope that this works out in JEE......And I feel, it's always better, we get confirmed.....I mean when u have a method under ur belt, why go for this guessing business?
Anyway, appearing tomorrow for olympiads?
