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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Find the sum of integers from 1 to 100 that are divisible...
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Find the sum of integers from 1 to 100 that are divisible by 3 or 5 ?

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There are total 33 numbers divisible by 3 within 1st 100 natural numbers. There are total 20 numbers divisible by 5 within 1st 100 natural numbers. There are total 6 numbers divisible by 15 within 1st 100 natural numbers. As multiples of 15 occur in both, we need to subtract it from addition...
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There are total 33 numbers divisible by 3 within 1st 100 natural numbers. There are total 20 numbers divisible by 5 within 1st 100 natural numbers. There are total 6 numbers divisible by 15 within 1st 100 natural numbers. As multiples of 15 occur in both, we need to subtract it from addition to cancel the repetition. sum of 'n' consecutive natural numbers starting from 1= n*(n+1)/2 Sum of numbers divisible by 3 = 3 * (1+2+3...33) = 3*33*34/2 = 99*17 use the trick = (100 - 1)*17 = 1700 - 17 = 1683 Similarly, Sum of numbers divisible by 5 = 5 * (1+2+3...20) = 5*20*21/2 = 50*21 = 1050 Sum of numbers divisible by 15 = 15 * (1+2+3...6) = 15*6*7/2 = 45 *7 = 315 Total sum = 1683+1050-315 = 2418 read less
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From 1 to 100, there are 33 multiples of 3 and 20 multiples of 5. Multiples of 3 are in A.P. and sum of the sequence is given by (n/2)(first term + last term), where n= no. of terms. For multiples of 3, n=33, first term= 3 and last term= 99. Similarly we calculate for multiples of 5, where n=20, 1st...
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From 1 to 100, there are 33 multiples of 3 and 20 multiples of 5. Multiples of 3 are in A.P. and sum of the sequence is given by (n/2)(first term + last term), where n= no. of terms. For multiples of 3, n=33, first term= 3 and last term= 99. Similarly we calculate for multiples of 5, where n=20, 1st term= 5 and last term= 100. However there are certain number that are multiples of both 3 & 5, so we need to subtract them and those are multiples of 15. For multiples of 15, n=6, 1st term= 15, last term=90. So using the formula mentioned above for A.P., we can solve. Required Sum = (Sum of multiples of 3) + (Sum of multiples of 5) - (Sum of multiples of 15) read less
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

=sum of integers divisible by 3 + Sum of integers divisible by 5 - Sum if integers divisible by 15. Each of these are simple AP Sums, so can be found easily.
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The integers from 1 to 100 which are divisible by 3=3,6,9.....99 this will form a AP with common difference 3 find the value of n using formula t(n)=a+(n-1)d u will get n=33 now find the sum of these integers for n=33 using sum formula in AP i.e S(n)=n/2 you will get S(n)=1683-------(1) Do the...
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The integers from 1 to 100 which are divisible by 3=3,6,9.....99 this will form a AP with common difference 3 find the value of n using formula t(n)=a+(n-1)d u will get n=33 now find the sum of these integers for n=33 using sum formula in AP i.e S(n)=n/2[2a+(n-1)d] you will get S(n)=1683-------(1) Do the above process for 5... The integers from 1 to 100 which are divisible by 5=5,10,15........100 this will form a AP with common difference 5 find the value of n using formula t(n)=a+(n-1)d u will get n=20 now find the sum of these integers for n=20 using sum formula in AP i.e S(n)=n/2[2a+(n-1)d] you will get S(n)=1050-----------------(2) Now find the integers which r divisible by both 3 and 5.. To find the integers take L.C.M of both number(3,5) i.e 15 So first No=15 Common Ratio=15 Last No=90 So N=6 Sum=315-------------------(3) Now the required Sum=1683+1050-315 =2418 read less
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we need to find sum of integers from 1 to 100 ,divisible by 3 or 5 so first find ( 3+6+ .......................... +99) then ( 5 +10+...................+100) now subtract the sum (15+............+90)...
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we need to find sum of integers from 1 to 100 ,divisible by 3 or 5 so first find ( 3+6+ .......................... +99) then ( 5 +10+...................+100) now subtract the sum (15+............+90) this is because we have to find n(A or B) means n(A+B)=n(A)+n(B)-n(A.B). hence answer is 2418 read less
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conceptual

Divisible by 3 form an a.p.to find sum of terms we have n/2*.repeat this for numbers divisible by 5 then u can get answer.
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Tutor - Maths and computer science

sum of integers from 1 to 100 that are divisible by 2 =n(n+1) =50*51 =2550 sum of integers from 1 to 100 that are divisible by 5 but not divisible by 2. are 5,15,.25,----------95 = 10/2 (5+95) = 500 The sum of integers from 1 to 100 that are divisible by 2 or 5 is=2550+500=3050 Ans.
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Specialization in Organic Chemistry. Teaching Chemistry has always been my passion.

= 3*(1+2+3+...+33) + 5*(1+2+3+...+20) - 15*(1+2+3+...+6)
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Online Tutor-Mathematics, Science, Competitive Exam Preparation

- Answer = 2418
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Ans is 2418
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