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Sequences and Series question

2021 · 26 Aug · Shift 2 · Q39
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Sequences and Series question

2021 · 26 Aug · Shift 2 · Q39

JEE MainMathematicsSequences and SeriesNumerical+4 / −1
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7975

  1. Divisibility rule for 11

For a 3-digit number abcabcabc, the number is divisible by 111111 if a−b+c=0, ±11.a-b+c=0,\ \pm 11.a−b+c=0, ±11. Since a,b,ca,b,ca,b,c are digits and the number is at most 500500500, we will list valid numbers directly.

  1. Allowed digits
  • The number is 3-digit, so the hundreds digit a≠0a\neq 0a=0.
  • Digit 111 is not allowed.
  • Number must be ≤500\le 500≤500, so a∈{2,3,4}a\in\{2,3,4\}a∈{2,3,4}.
  • Other digits can be from {0,2,3,4,5,6,7,8,9}\{0,2,3,4,5,6,7,8,9\}{0,2,3,4,5,6,7,8,9}.
  1. Form of 3-digit multiples of 11

A 3-digit multiple of 111111 has the form: 110,121,132,143,154,165,176,187,198,209,220,231,242,253,264,275,286,297,308,319,330,341,352,363,374,385,396,407,418,429,440,451,462,473,484,495110,121,132,143,154,165,176,187,198,209,220,231,242,253,264,275,286,297,308,319,330,341,352,363,374,385,396,407,418,429,440,451,462,473,484,495110,121,132,143,154,165,176,187,198,209,220,231,242,253,264,275,286,297,308,319,330,341,352,363,374,385,396,407,418,429,440,451,462,473,484,495 up to 500500500.

  1. Remove numbers containing digit 111

From the above list, exclude all numbers having digit 111: 121,143,154,165,176,187,198,319,341,418,451121,143,154,165,176,187,198,319,341,418,451121,143,154,165,176,187,198,319,341,418,451 Also 110110110 contains digit 111, so exclude it too.

Thus the valid numbers are: 209,220,231,242,253,264,275,286,297,308,330,352,363,374,385,396,407,429,440,462,473,484,495209,220,231,242,253,264,275,286,297,308,330,352,363,374,385,396,407,429,440,462,473,484,495209,220,231,242,253,264,275,286,297,308,330,352,363,374,385,396,407,429,440,462,473,484,495

  1. Add them

Let us sum carefully: [ \begin{aligned} &209+220+231+242+253+264+275+286+297 \ &= 2277 \end{aligned} ]

[ \begin{aligned} &308+330+352+363+374+385+396+407+429+440+462+473+484+495 \ &= 5698 \end{aligned} ]

So total sum is 2277+5698=7975.2277+5698=7975.2277+5698=7975.

  1. Final answer

The required sum is 7975.\boxed{7975}.7975​.

  1. Comparison with stored answer

Stored correct answer is 774477447744, but the correct computed sum is 797579757975. Hence, I do not agree with the stored answer.

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