Math, asked by priyankaaggarwal024, 11 months ago

If a is an integer and b=3a+7 then which of the following cannot be a divisor of b? (A)4 (B) 9 (C) 11 (D)13

Answers

Answered by amitnrw
1

9 can not be a divisor of  b

Step-by-step explanation:

b=3a+7

=>b = 3a + 6  + 1

=> b = 3(a + 2)  + 1

=> b = 3k + 1

=> 3 can not be a divisor of b

Hence 9 can not be a divisor of  b

b = 3a + 7

a = 3  => b = 16  ( 4 is divisor)

a = 5 => b = 22 (11 is divisor)

a = 2 => b = 13  ( 13 is divisor)

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Answered by kunisir
0

Answer:

Given a is an integer .

And b= 3a+7

Now b can be put into all these integral forms m (k) + n

(i) b= 3a+7 =(3a +3)+4 = 3(a+1) +4 = 3k+4

(ii) b= 3a+7 =(3a +18) -11 =3 (a+6) -11=3k-11

(iii) b=3a+7=(3a-6)+13 =3(a-2) +13=3k=13

(iv) b= 3a+7=(3a +6) +1 =3(a+2)=1

Putting an integer 0 for the value of k,we find b to be  a divisor of 4, 11 and 13, but not 9.

So Answer is (B) 9

Step-by-step explanation:

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