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Explore the Back-StoryTry this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

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

Divisibility

Equations

Integer

Answer: is 64.

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

{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

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)

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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

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

Divisibility

Equations

Integer

Answer: is 64.

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

{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

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)

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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