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.
If x=\left(16^{3}+17^{3}+18^{3}+19^{3} \right), then x divided by 70 leaves a remainder of what?
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8 Answers
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.
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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?
