Number distribution riddle
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Answered
N*(N+1)/2 sum of n positive integers starting from 1.
so each group sum should be N*(N+1)/2*5
99*100/10 = 990
Best answer
Total sum of numbers is equal to 4950.
This implies that the sum of numbers in each row should be equal to 990.
group 1 = 1-9 and 90-99
group 2 = 10-19 and 80-89
group 3 = 20-29 and 70 to 79
group 4 = 30-39 and 60-69
group 5 = 40-59
Your Answer
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