SUM OF SEQUENCE

Number Theory Level 1

Obtain the sum of all integers in the first 1000 integer which are neither divisible by 5 nor by 2

200,000 200,654 600,000 200,009

Muti Ur Rehman Kamboh by Muti Ur Rehman Kamboh , 27, Pakistan

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2 solutions

the number which are not divided on 5 nor 2 are 1,3,7,9,11,13,17,19,21,23,27,29....991,993,997,999 required sum is =1+3+7+9+11+13+17+19+21+23+27+29+.......+991+993+997+999 now (1+11+21+.....+991)+(3+13+23+.....+993)+(7+17+27+.....+997)+(9+19+29+...+999) in each group the number of term are same let xn = a + d(n-1) xn =991 d=11 1=10 a=1 991=1+10(n 1) n=100 thus required sum is S = (n/2) × (2a + (n-1)d) now put the values S=[100/2(2+(99)10]+[100/2(2(3)+(99)10]+[100/2(2(7)+(99)10]+[100/2(2(9)+(99)10] =50[2+990+6+990+14+990+18+990] S=200,000

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there are [1000/5]=200 numbers divisible by 5. Sum of them is A=5 200 201/2=100500

there are [1000/2]=500 numbers devisible by 2. sum of them is B=2 500 501/2=250500

there are [1000/10]=100 numbers devisible by both 2 and 5. Sum of them is C=10 100 101/2=50500

sum of all numbers devisible by 2 or 5 would be D=A+B-C=300500

sum of all numbers between 1 and 1000 is E=1000*1001/2=500500

the final answer would be E-D=200,000

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