Call the greatest number $n$ and the common remainder $x$, so our problem is $$\begin{align} x&\equiv 43\pmod{n}\\ x&\equiv 91\pmod{n}\\ x&\equiv 183\!\!\!\pmod{n} \end{align}\qquad$$ By general CRT theory this system is solvable iff pairwise solvable, i.e. iff $$\begin{align} &n\mid 91\!-\!43,\, 183\!-\!91,\, 184\!-\!43\\ \iff \ \ &n\mid 48,92,140\\ \iff\ \ &n\mid \gcd(48,92,140) = 4\end{align}\ \ $$ where the final arrow is by the gcd Universal Property. Open in App Suggest Corrections 21
Find the greatest number that will divide 43, 91 and 183 so as to leav [#permalink]
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Question Stats: Hide Show timer StatisticsFind the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.A. 4B. 7C. 9D. 13 E. 24 _________________
Originally posted by LMP on 09 Aug 2018, 00:35. Renamed the topic.
Find the greatest number that will divide 43, 91 and 183 so as to leav [#permalink]
Bulusuchaitanya wrote: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.A. 4B. 7C. 9D. 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 AE) 2443 divided by 24 equals 1 with remainder 1991 divided by 24 equals 3 with remainder 19183 divided by 24 equals 7 with remainder 15 We need the SAME remainder each time - ELIMINATE ED) 1343 divided by 13 equals 3 with remainder 491 divided by 13 equals 7 with remainder 0We need the SAME remainder each time - ELIMINATE DC) 943 divided by 9 equals 4 with remainder 791 divided by 9 equals 10 with remainder 1We need the SAME remainder each time - ELIMINATE CB) 743 divided by 7 equals 6 with remainder 191 divided by 7 equals 13 with remainder 0We need the SAME remainder each time - ELIMINATE BBy the process of elimination, the correct answer is ARELATED VIDEO _________________
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Re: Find the greatest number that will divide 43, 91 and 183 so as to leav [#permalink]
Bulusuchaitanya wrote: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.A. 4B. 7C. 9D. 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] 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]
Bulusuchaitanya wrote: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.A. 4B. 7C. 9D. 13 E 24 91=pn+r43=qn+rsubtracting,48=n(p-q)n must be 4 or 24only 4 leaves same remainder with all three dividendsA
Re: Find the greatest number that will divide 43, 91 and 183 so as to leav [#permalink]
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]
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)=140It is obvious that q3>q2>q1For 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 [#permalink]
Bulusuchaitanya wrote: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.A. 4B. 7C. 9D. 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 - AnswerIf 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] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Find the greatest number that will divide 43, 91 and 183 so as to leav [#permalink] 20 May 2021, 06:14 |
What is the greatest number that will divide 43 91 and 184 so as to leave the same remainder in each case?
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