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A
arundhatip4374
January 11, 2023
Question

Want to display numbers from 105

  • January 11, 2023
  • 13 replies
  • 0 views

Hi Everyone,

i am trying to display numbers from 105 but the output should be : {105,110,115,120,,125,130,135,140,145,150,151,152,153,154,155,156,157,158,159,160,165,170,175,180,185,190,195,200},

i have tried but after 105 iam getting 115,Here sharing my code 

a!localVariables(
local!data: 101 + enumerate(100),
index(
local!data,
where(like(touniformstring(local!data), "*5*"))
)
)

Thank You In Advance

    13 replies

    D
    deepakg681722
    January 11, 2023

    Could you please try the below code.

    a!localVariables(
  local!data: 101 + enumerate(100),
  index(
    local!data,
    (enumerate(count(local!data)/5)+1)*5,
    {}
  )
)

    A
    arundhatip4374Author
    January 11, 2023

    its only showing those are divided by 5 i want to show that but as well after 150 i want to show 151,152,153,154,155,156,157,158,159 like this then after 160 again 165,170

    D
    deepakg681722
    January 11, 2023

    Something like below?

    a!localVariables(
  local!data: 101 + enumerate(100),
  local!indexesContains5: wherecontains(true(), find(5, local!data, 1) > 1),
  local!multipleOf5: (enumerate(count(local!data) / 5) + 1) * 5,
  sortintegerarray(
    index(
      local!data,
      union(
        local!multipleOf5,
        local!indexesContains5
      ),
      {}
    ),
    false()
  )
)

    S
    sreesuryanagag0001
    January 11, 2023

    a!localVariables(

    local!test:a!forEach(

    items: 101+enumerate(100),

    expression:if(mod(fv!item,5)=0,fv!item,{}) ),

    local!test

    )

    R
    ramakrishnaa786409
    January 11, 2023

    a!localVariables(
  local!data:a!forEach(
    items: enumerate(100)+101,
    expression: {if(mod(fv!index,5),{},fv!item),if(find(5,fv!item),fv!item,{})}
  ),
 union(local!data,local!data)
)

      R
      rakeshg3074
      January 11, 2023

      Hi [mention:def63b4d161e40ea8ebfb4b690b0b18c:e9ed411860ed4f2ba0265705b8793d05]

      Once try it with this..

      a!localVariables(
      local!data: a!forEach(
      items: enumerate(101)+100,
      expression: append(if(find(5,fv!item),fv!item,{}),if(mod(fv!item,5),{},fv!item)),
      ),
      local!uniqueData:union(local!data,local!data),
      local!uniqueData
      )

      A
      adeleb9274
      January 16, 2023

      i just saw my work has the same error. tks 

      papa's games

      G
      gailp042440
      June 16, 2025

      Did you manage to solve it?

      awesome tanks 2

      harshas2775
      harshas2775
      June 16, 2025

      Looks like by the 'verified' answer above they were able to solve it.

      stefanhelzle0001
      stefanhelzle0001
      January 31, 2023

      Nice challenge, but for what purpose? Below my approach.

      {
  105 + enumerate(9) * 5,
  150 + enumerate(11),
  160 + enumerate(9) * 5,
}

      mikes0011
      mikes0011
      Brainy
      January 31, 2023
      Nice challenge

      indeed.

      i was, also, wondering if this could be reduced to something more elegant/simple and I just came up with this:

      (enumerate(20) + 21) * 5

      Though of course a lot depends on unfilled assumptions - how long of a list do we need?  Are we assuming it ends at "200" or is that just an example ending point?

      sureshd0004
      sureshd0004
      June 16, 2025

      Old post but interesting issue, hence posting this code for anyone else interested in it. 

      a!localVariables(
  local!startNUmber: 105,
  local!endNUmber: 200,
  local!diffrence: 5,
  local!diff: (
    (local!endNUmber - local!startNUmber) / local!diffrence
  ) + 1,
  {
    local!startNUmber + enumerate(local!diff) * local!diffrence
  }
)

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