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Find the greatest number that will divide 43, 91 and 183 so as to leav
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02 Nov 2018, 21:59
Gagoosh wrote:
alok019 wrote:
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.
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 !! _________________
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav
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03 Nov 2018, 00:10
2
Bulusuchaitanya wrote:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
A. 4 B. 7 C. 9 D. 13 E. 24
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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