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

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

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

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

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
...

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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