Rounding up 5492 to 8000 “FAILURE”

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  • #17041
    ARTURO CHAVEZ MOGOLLON
    Participant
    • Topics: 3
    • Replies: 4

    Hello. First of all, I would like to thank you for your kind help and support, dear Of-Pad.

    I have been checking if my question had already been asked in here and, apparently, the answer is “no” hehe.

    In your book “Mental Math”, Kindle version, page 80, example #2, instead of rounding 5492 to 5500, I am trying to do it to 8000 so I can sum more easily with 8739 and get 16 739. BUT, … I bump into a tiny problem.

    5492 = 8000 – #

    So, as far as I understood, the number of digits of the answer should be equivalent to the number of zeros of the target number (8000 in this case) . AND, as far as I understood, if both numbers (5492 & 8000 in this case) have only 4 digits, then the answer of the rounding up calculation shouldn’t be 4 digits, but instead “3”; so both understandings of mine make the most sense…unless I am wrong somewhere, and…if that’s the case, I apologise and beg for your correction of my mistake.

    SO, if I apply the aforementioned understandings to round up 5492 to 8000…I have that the answer should be “508” BUT

    5492 = 8000 – 508  (Wrong) …so how can I work out the missing “2” at the beginning of 508?

    Thank you soooo much in advance, Of Pád. You’re great!! I am learning a lot because of you and I am having quite a good time using my head during this pandemic thanks to you 🙂

    #17042
    ARTURO CHAVEZ MOGOLLON
    Participant
    • Topics: 3
    • Replies: 4

    I think I might have found something… I don’t know, tell me what you think, anyway

    For example:

    6529 = 8000 – ????

    I first round up the last 3 digits of 6529  (= *529), which is the number of zeros of 8000 (3 zeros, so 3 digits)

    9 + 1  = 10 ;   1

    2 + 7 = 9    ;   7

    5 + 4 = 9  ;    4

     

    Then, for 6 (of 6529) and 8 (of 8000), I think I have found a rule.  If the numbers differ in 1, then it’s “0”; if the numbers differ in 2, then it’s “1”; if the numbers differ in 3, then it’s “2”and so on This is what I mean:

    6529 = 7000 – 0471

    6529 = 8000 – 1471

    6529 = 9000 – 2471

    6529 = 9000 – 3471

     

    Furthermore, this rule applies to the problem of my question on the post above. The one of   “ 5492 = 8000 – #    ”

    Between 5 (of 5492) and 8 (of 8000), they differ in “3” so it would be “2” the first number of the answer. So,

    5492 = 8000 – 2508

     

    Tell me what do you think and if you have another method for this 🙂

     

    #17537
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 73

    Sorry for the late reply. I was on vacation but looks like you figured it out.

    The method in the book is just to round up to the nearest multiple of 10, 100 or 1000.

    In your question 5492 = 8000 – ____, the nearest multiple of 5492 is actually 6000 not 8000.

    So 5492 = 6000 – 508.

    If you want to round up to the next multiple of 1000 which is 7000 then you add 1000 to 508 to get 1508.

    If you want to round up to 8000 then you need to add 2 x 1000 = 2000.

    You can subtract the first digit of both numbers and then subtract the difference by one.

    In 5492 = 8000 – _____

    8 – 5 – 1 = 2

    9 – 4 = 5

    9 – 9 = 0

    10 – 2 = 8

    So the rounded up number is 2508

    5492 = 8000 – 2508

    Hope this provides more clarity.

    #17540
    ARTURO CHAVEZ MOGOLLON
    Participant
    • Topics: 3
    • Replies: 4

    Hi Abhishek!

    Thank you soo much for your reply! Now I have a quite better understanding about this particular aspect of Mental Math. 🙂 You’re quite a genius, honestly!

    I have already posted my review on your book on Amazon.co.uk and rated it as well, and now I am making a pause on your book “Mental Math” so I can read the other book of yours called “The Math Palace Technique”. I love your work!

    See you soon!

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