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Bulusuchaitanya
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) 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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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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Bulusuchaitanya
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
A
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alok019
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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Bulusuchaitanya
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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We need to find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A. 4
43, 91, 183 each when divided by 4 gives 3 remainder => POSSIBLE

B. 7
43 when divided by 7 gives 1 remainder
91 when divided by 7 gives 0 remainder => NOT POSSIBLE

C. 9
Sum of all digits of 43 = 4+3 = 7 => Remainder of 43 by 9 = Remainder of 7 by 9 = 7
Sum of all digits of 91 = 9+1 = 10 => Remainder of 91 by 9 = Remainder of 10 by 9 = 1 => NOT POSSIBLE

D. 13
43 when divided by 13 gives 4 remainder
91 when divided by 13 gives 0 remainder => NOT POSSIBLE

E. 24
43 when divided by 24 gives 19 remainder
91 when divided by 24 gives 19 remainder
183 when divided by 24 gives 15 remainder=> NOT POSSIBLE

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Divisibility Rules


Watch the following video to learn the Basics of Remainders

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