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I found the following method to find one's age. It is working exactly for my case. I like to understand and solve this puzzle. If this is wrong forum, my sincere apologies. Please guide me solve this.

  1. Take last digit of your mobile number
  2. Multiply it by 2
  3. Add 5 with the above answer
  4. Multiply the answer with 50
  5. Add 1764 with the answer
  6. Subtract the your birth year from the result in part 5 (example 1985, 1973, etc.)

Now you will get 3 digit number. The first one is last digit of your mobile number, last two digits are your age.

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    $\begingroup$ What year is meant in 6., which is the only point where your age might sneak in. $\endgroup$
    – mvw
    Aug 1, 2014 at 17:40
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    $\begingroup$ Should step 3 be "Add 1 to the above answer"? $\endgroup$
    – Daniel R
    Aug 1, 2014 at 17:41
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    $\begingroup$ Then I'm puzzled. How does it work for your age? Or should step 5 be changed to "add 1764 to the answer"? $\endgroup$
    – Daniel R
    Aug 1, 2014 at 17:55
  • 3
    $\begingroup$ Old trick. Back in the day, when mobile devices didn't exist, we had to choose a number between 0 and 9. How how times have changed... $\endgroup$ Aug 1, 2014 at 23:44
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    $\begingroup$ Won't work for anyone above the age of 99. $\endgroup$
    – G. Bach
    Aug 2, 2014 at 20:43

3 Answers 3

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  1. Let $n$ be the last digit of your mobile number.
  2. $2n$
  3. $2n+5$
  4. $50(2n+5)=100n+250$
  5. $100n+250+1764=100n+2014$ (fixed by Daniel R)
  6. Let $y$ be your year of birth. Then we have $100n+(2014-y)$.

Since $2014-y$ is your age (assumed to be $<100$), the hundreds digit is $n$ and the last two digits are your age.

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    $\begingroup$ And this assume this trick is performed in the year 2014. $\endgroup$ Aug 2, 2014 at 1:16
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    $\begingroup$ " 2014−y is your age" -- except that it isn't, since I haven't had my birthday yet this year. But, you know, maths is nearly psychic. $\endgroup$ Aug 2, 2014 at 11:42
  • $\begingroup$ Since $n$ is single digit choosen it will get in hundreds digit. $\endgroup$
    – Fawad
    Sep 28, 2016 at 6:40
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That's elementary. If $d$ is the last digit and $y$ your birth year, you are computing

$$(2d+1)50+1964-y=100d+2014-y.$$

The first term is the digit shifted two positions to the left (times $100$); the other two terms compute your age.

You'll have to increment the constant $1964$ every year, and the trick won't work for centenarians.

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  • $\begingroup$ This answer is also correct. Thanks @Yves Daoust $\endgroup$
    – rajakvk
    Aug 1, 2014 at 18:10
  • $\begingroup$ It's obviously correct, but how are you getting (2d+1)50 + 1964 from what is stated? I get that you can rearrange (2d+5)50 + 1764 that way, but that seems like extra work. So I assume you get it some other way. $\endgroup$
    – trlkly
    Aug 2, 2014 at 12:08
  • $\begingroup$ @trlkly: the original text was different. $\endgroup$
    – user65203
    Aug 3, 2014 at 9:20
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$50(2x + 1) + 1964 - y = 100x + (2014 - y).$

If you were born before August 1st in year y, then (2014 - y) is your age. If you are under 100, then it's a 2-digit number. So the right-hand side of the equation is just a three digit number whose first digit is $x$ (doesn't matter what one-digit number $x$ we started with) and whose last two digits are your age.

edit: fixed lhs as per comments

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  • $\begingroup$ This is not true as written; you need $(2x + 1)$ for this to work $\endgroup$ Aug 1, 2014 at 17:45
  • $\begingroup$ It's obviously correct, but how are you getting 50(2x+1) + 1964 from what is stated? I get that you can rearrange 50(2x+5) + 1764 that way, but that seems like extra work. So I assume you get it some other way. $\endgroup$
    – trlkly
    Aug 2, 2014 at 12:10
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    $\begingroup$ the question appears to have been edited to change the number since i made my answer. $\endgroup$
    – hunter
    Aug 2, 2014 at 12:15
  • $\begingroup$ @hunter Thanks. Next time, though, I would prefer it if you use the @username notation when you leave a reply to me. That way, the reply appears in my inbox. I only happened to see that this question in the "Hot Questions" part or I would have missed your reply. $\endgroup$
    – trlkly
    Aug 3, 2014 at 3:45
  • $\begingroup$ hunter, my bad, yes I edited question. @trlkly sorry for the confusion. $\endgroup$
    – rajakvk
    Aug 5, 2014 at 12:27

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