When 100 is to be successively divided by 6, 3, 4, first divide 100 by 6. Then…
2025
When 100 is to be successively divided by 6, 3, 4, first divide 100 by 6. Then divide the quotient 16 by 3. Then divide the quotient 5 by 4. A number when successively divided by 5, 3, 2 gives the remainder of 0, 2 and 1 respectively in that order. What will be the remainders when the same number is divided successively by 2, 3 and 5 in that order?
- A.
1, 0, 4
- B.
4, 3, 2
- C.
4, 1, 2
- D.
2, 1, 3
Attempted by 8 students.
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Correct answer: A
Concept: In successive division, a number N divided by the first divisor d₁ leaves remainder r₁ and quotient q₁, so N = d₁·q₁ + r₁. That quotient q₁ is then divided by the next divisor d₂ to leave remainder r₂ and quotient q₂, so q₁ = d₂·q₂ + r₂, and so on. Reconstructing N works by substituting each equation back into the previous one, starting from the smallest valid (non-negative) quotient at the last stage.
Let the number be N. Successively dividing by 5, 3, 2 (in that order) leaves remainders 0, 2, 1, so: N = 5X + 0, X = 3Y + 2, Y = 2Z + 1.
Take the smallest non-negative quotient at the last stage, Z = 0 (a quotient of 0 is valid). Then Y = 2(0) + 1 = 1.
Substitute back: X = 3(1) + 2 = 5.
Substitute back: N = 5(5) + 0 = 25.
Now divide N = 25 successively by 2, 3, 5 (the required new order): 25 ÷ 2 gives quotient 12, remainder 1.
12 ÷ 3 gives quotient 4, remainder 0.
4 ÷ 5 gives quotient 0, remainder 4.
Cross-check: any number of the form 25 + 30k (since 2×3×5 = 30) satisfies the original successive-division data - e.g. N = 55 (k = 1) also works - and because 30 is divisible by 2, 3 and 5, every such family member gives the identical remainder sequence, so the answer does not depend on which family member is picked. As an independent check, dividing 25 successively by 5, 3, 2 forward reproduces the given remainders 0, 2, 1, confirming N = 25 is consistent with the original condition.
Result: the remainders are 1, 0, 4, matching the option 1, 0, 4.