Q.

If  36Cr+1k23=6×35Cr then the number of ordered pairs (r, k) where k  Z, is 

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a

2

b

6

c

3

d

4

answer is D.

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Detailed Solution

We have,

 36Cr+1k23=6×35Cr36r+135Crk23=35Cr×6k23=r+16k2=3+r+16

Since k2 is an integer and r takes values from Oto 35. Therefore,

r=5 or 35

Putting r=5 in (i), we obtain

k2=3+1=4k=±2

Putting r=35 in (ii), we obtain

k2=3+6=9k=±3

Hence, the total number of order pairs (r, k) is 4. 

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If  36Cr+1k2−3=6×35Cr then the number of ordered pairs (r, k) where k ∈ Z, is