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?
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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.]

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