Latest Accenture Aptitude Question SOLUTION: Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.

• Placement Papers
• /
• Accenture
• /
• Numerical Ability
• /
• LCM and HCF

Accenture Question Papers: Accenture Interview Experience(7) ACCENTURE Company Profile

(#M40110960) ACCENTURE Numerical Ability question Keep an EYE

Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.

Asked In Accenture chandreyee chowdhury (9 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (22)   (12) Submit Your Solution LCM and HCF

Read Solution (5) :

Yasaswi  (9 years ago)

(26-22,39-35,52-48,65-61) = (4,4,4,4)
LCM of (26,39,52,65) =780
Req. number= 780-4 =776

Is this solution Helpfull? Yes (9) | No (1)  

Vinay Kumar singh  (8 years ago)

since the common difference between remainder and divisor is constant i.e 26-22=4,39-35=4,52-48=4,65-61=4
so the no will be =L.C.M of(26,39,52,65)-4
=780-4
=776

Is this solution Helpfull? Yes (2) | No   

B.Naga Sivani  (8 years ago)

776 is ans

Is this solution Helpfull? Yes | No   

tamilmani  (8 years ago)

the answer is 776
26-22=4
39-35=4
52-48=4
65-61=4
LCM(26,39,52,65)=780
780-4=776

Is this solution Helpfull? Yes | No   

Divya Batra  (6 years ago)

Lets take LCM of all four divisors= 780 is the LCM common multiple of all numbers. so, when these numbers divide 780 by themselves, the remainder is the divisor itself., means,
780%26=26
same goes for other numbers, so we need a number which leaves the given remainders in the questions, so we subtract 26-22=4, from 780, gives 776.

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.