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Find the greatest number that will divide 43, 91 and 183 so as to leav

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Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post Updated on: 10 Aug 2018, 00:22
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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

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Originally posted by LMP on 09 Aug 2018, 01:35.
Last edited by Bunuel on 10 Aug 2018, 00:22, edited 1 time in total.
Renamed the topic.
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 09 Aug 2018, 03:06
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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



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

Ans. (A)
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 09 Aug 2018, 06:47
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
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 09 Aug 2018, 07:54
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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


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
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 28 Aug 2018, 13:44
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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


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) 41
43 divided by 41 equals 1 with remainder 2
91 divided by 41 equals 2 with remainder 9
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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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 31 Aug 2018, 09:41
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.
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Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 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.
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav  [#permalink]

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New post 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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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav   [#permalink] 03 Nov 2018, 00:10
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