Discussion
If a set of toys is arranged in rows of 20 each, 15 toys are left over; in rows of 25 each, 20 are left over; in rows of 38 each, 33 are left over; and in rows of 40 each, 35 are left over. What is the minimum number of toys?
Answer: D
In every case the shortfall is the same:
\[ 20 - 15 = 25 - 20 = 38 - 33 = 40 - 35 = 5 \]
So \( N + 5 \) is divisible by all four numbers.
\[ \text{LCM}(20, 25, 38, 40) = 2^{3} \times 5^{2} \times 19 = 3800 \]
\[ N = 3800 - 5 = 3795 \]
Spotting the common shortfall is what turns four separate congruences into a single LCM.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development