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Natural Numbers Problem | PRMO 2019 | Question 30

Try this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

Natural numbers Problem - PRMO 2019


Let E denote the set of all natural number n such that \(3< n<100\) and the set {1,2,3,...,n} can be partitioned in to 3 subsets with equal sums. Find the number of elements of E.

  • is 107
  • is 64
  • is 840
  • cannot be determined from the given information

Key Concepts


Divisibility

Equations

Integer

Check the Answer


Answer: is 64.

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

Try with Hints


First hint

{1,2,...,n}

This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3

\(\frac{n(n+1)}{2}\) is divisible by 3.

or, n of form 3k, 3k+2

or, n=6k,6k+2,6k+3, 6k+5

Second Hint

case I n=6k, we group numbers in bundles of 6 for each bundle 1,2,3,4,5,6(16,25,34)

case II n=6k+2 then we club last bundle of 8 numbers rest can be partitioned and those eight numbers can be done 1,2,3,4,5,6,7,8 (1236,48)

case III n=6k+3 we club last nine number and rest can be partitioned 1,2,3,4,5,6,7,8,9 (12345,69,78)

Final Step

case IV 6k+5 we take last five numbers, rest can be aprtitioned 1,2,3,4,,5(14,25,5)

Hence we select any number of form 6k(16), 6k+2(16), 6K+3(16), 6K+5(16)

or, total=64 numbers.

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Try this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

Natural numbers Problem - PRMO 2019


Let E denote the set of all natural number n such that \(3< n<100\) and the set {1,2,3,...,n} can be partitioned in to 3 subsets with equal sums. Find the number of elements of E.

  • is 107
  • is 64
  • is 840
  • cannot be determined from the given information

Key Concepts


Divisibility

Equations

Integer

Check the Answer


Answer: is 64.

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

Try with Hints


First hint

{1,2,...,n}

This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3

\(\frac{n(n+1)}{2}\) is divisible by 3.

or, n of form 3k, 3k+2

or, n=6k,6k+2,6k+3, 6k+5

Second Hint

case I n=6k, we group numbers in bundles of 6 for each bundle 1,2,3,4,5,6(16,25,34)

case II n=6k+2 then we club last bundle of 8 numbers rest can be partitioned and those eight numbers can be done 1,2,3,4,5,6,7,8 (1236,48)

case III n=6k+3 we club last nine number and rest can be partitioned 1,2,3,4,5,6,7,8,9 (12345,69,78)

Final Step

case IV 6k+5 we take last five numbers, rest can be aprtitioned 1,2,3,4,,5(14,25,5)

Hence we select any number of form 6k(16), 6k+2(16), 6K+3(16), 6K+5(16)

or, total=64 numbers.

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