JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 17)
The greatest positive integer k, for which 49k + 1 is a factor of the sum
49125 + 49124 + ..... + 492 + 49 + 1, is:
49125 + 49124 + ..... + 492 + 49 + 1, is:
32
60
63
65
Explanation
1 + 49 + 492
+ ..... + 49125
sum of G.P. = $${{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}$$
= $${{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}$$
Also 4963 - 1
= (1 + 48)63 - 1
= [63C0$$ \times $$1 + 63C1$$ \times $$ 48 + 63C2$$ \times $$ (48)2 + .... ] - 1
= [1 + 48$$\lambda $$] - 1 = 48$$\lambda $$
So $${{\left( {{{49}^{63}} - 1} \right)} \over {48}}$$ = integer
$$ \therefore $$ 4963 + 1 is a factor.
So k = 63.
sum of G.P. = $${{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}$$
= $${{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}$$
Also 4963 - 1
= (1 + 48)63 - 1
= [63C0$$ \times $$1 + 63C1$$ \times $$ 48 + 63C2$$ \times $$ (48)2 + .... ] - 1
= [1 + 48$$\lambda $$] - 1 = 48$$\lambda $$
So $${{\left( {{{49}^{63}} - 1} \right)} \over {48}}$$ = integer
$$ \therefore $$ 4963 + 1 is a factor.
So k = 63.
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