When the three numbers Q, (Q+384) and (Q+528) are divided by the whole number, N, the remainder is the same. What is the greatest possible value of N?

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Because they have the same remainder the difference between the numbers Q+528 and Q+384=144 must be an exact multiple of N. So we can write N∆₁=Q+528-Q-384=144, where ∆₁ is the difference between the multipliers.

Also, Q+384-Q=384 must be an exact multiple of N, so we can write N∆₂=384, where ∆₂ is the difference between the multipliers. N∆₂/N∆₁=384/144=8/3.

384/8=144/3=48, so 48 is the HCF between 384 and 144. Therefore 48 is the largest value for N.


[Q=48a+r, Q+384=48b+r, Q+528=48c+r, where r<48 is the common remainder, and a, b and c produce multiples of 48: 48a, 48b, 48c, and ∆₁=c-b=144/48=3, ∆₂=b-a=384/48=8.

EXAMPLE: Put a=1, r=35. Then b=9 and c=12. So Q=83, Q+384=467, Q+528=611. Dividing each of the three numbers: 83, 467, 611, by 48 produces the same remainder.]

by Top Rated User (1.0m points)

Related questions

1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,184 questions
97,329 answers
24,564 users