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.
Divisibility
Equations
Integer
But try the problem first...
Answer: is 64.
PRMO, 2019, Question 30
Elementary Number Theory by David Burton
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.
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.
Divisibility
Equations
Integer
But try the problem first...
Answer: is 64.
PRMO, 2019, Question 30
Elementary Number Theory by David Burton
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.
