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Subhash B

· started a discussion

· 1 Months ago

Either question is wrongly printed or sol is wrong (Question: 107412.0 )

Sol:(if question is correct)
(2 pow 39 ) /39 = ((39-37) pow 39)/39 = (39.x -37)/39

(where x in 39.x is the divisible part other than last term in binomial expansion of ((39-37)pow 39).

so (39(x-1)+39-37)/39= (39(x-1)/39) + (2/39)
clearly remainder would be 2.

Question:
The remainder when 239 is divided by 39 is:
Options:
A) 0
B) 2
C) 8
D) 1
Solution:
Ans: (c)

\(\cfrac{2^{39}}{39}\ = \cfrac{(2^6)^6\times 2^3}{39}\ = \cfrac{(25)^6\times 2^3}{39}\)

\(\cfrac{(625)^3\times 2^3}{39} \ =\cfrac{(1)^3\times 8}{39} \ = \cfrac{8}{39}\)

Hence 8 will be remainder.

Subhash B

· commented

· 1 Months ago

Yet to be approved!
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