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Q.

Let the number (22)2022+(2022)22 leave the remainder α when divided by 3 and β when divided by 7. Then 
S1:(α2+β2)=5  
S2:(α+β)2=9 
S3:(αβ)2=1 
S4:(αβ)=2 
Then, the number of statements which are true is

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a

2

b

1

c

3

d

4

answer is B.

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Detailed Solution

(22)2022+(2022)22

Divided by 3, 

(21+1)2022+(2022)22 [  2022=3×67.4]=3k+1 (Remainder =α+1

Divided by 7

(21+1)2022+(20231)22

7k+2(β=2)  [   As2023=7×289]

So, α2+β2=5

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