This question lends itself to testing the answer choices
Since we're looking for the greatest number, we'll start at E and work our way to A

E) 24
43 divided by 24 equals 1 with remainder 19
91 divided by 24 equals 3 with remainder 19
183 divided by 24 equals 7 with remainder 15
We need the SAME remainder each time - ELIMINATE E

D) 13
43 divided by 13 equals 3 with remainder 4
91 divided by 13 equals 7 with remainder 0
We need the SAME remainder each time - ELIMINATE D

C) 9
43 divided by 9 equals 4 with remainder 7
91 divided by 9 equals 10 with remainder 1
We need the SAME remainder each time - ELIMINATE C

B) 7
43 divided by 7 equals 6 with remainder 1
91 divided by 7 equals 13 with remainder 0
We need the SAME remainder each time - ELIMINATE B

By the process of elimination, the correct answer is A

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Greatest number=HCF(91-43, 183-91, 183-43)=HCF(48, 92, 140)=4.

Ans. (A)
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The answer is 4 and how it is 4 is below,

We can represent any integer number in the form of: D*q + r.
Where D is divisor, q is quotient, r is remainder.

so each number can be written accordingly:
43 = D*q1 + r1;
91 = D*q2 + r2;
183 = D*q3 + r3;

r1,r2 & r3 will be same in above three equations according to the question.
D is the value that we want to find out. which should be greatest.

On solving three equations we get:

D*(q2-q1)= (91-43)=48
D*(q3-q2)= (183-91)=92
D*(q3-q1)= (183-43)=140
It is obvious that q3>q2>q1
For the greatest value of D that divide each equation we take the HCF of 48,92,140

THEREFORE ANSWER IS 4.
Please give kudos !!

You are looking for the greatest divisor so you are looking for HCF. Say it is H. Say the common remainder is R.

43 = Ha + R ... (I)
91 = Hb + R ... (II)
183 = Hc + R ... (III)

(II) - (I)
48 = H(b - a)

(III) - (I)
140 = H(c - a)

So H has to be a factor of 48 (= 2^4*3) as well as 140 (= 2^2 * 5 * 7). So highest value of H can be 4 as of now. Considering equation (III) - (II) we might get that it can be 2 only, we don't know yet. But note that the options have only 4 and hence answer (A)

Alternatively, try out the options.
If 4 is the divisor, remainders are 3, 3, 3 - Answer
If 7 is the divisor, remainders are 1, 0 - Not the answer
...
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91=pn+r
43=qn+r
subtracting,
48=n(p-q)
n must be 4 or 24
only 4 leaves same remainder with all three dividends
A

required no.=h.c.f of(91-43),(183-91) and (183-43)
=h.C.F OF 48,92 and 140=4
hence A option is correct

could u please explain this mthod in detail? would much appreciate the help.

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