15
$\begingroup$

All 5 numbers are entered in an ascending order. These numbers are in an arithmetic progression. The value of the fifth number, i.e. the largest number, is less than 99. While adding these numbers, two plus(+) symbols were accidentally not entered due to the bad calculator, giving the sum as 71157.
And while finding the product of these numbers, two multiplication symbols were not typed, resulting in the mistaken product as 11698071.
In the places where + were entered correctly in the first operation, multiplication symbols were missing there in second operation and vice versa.

Find those 5 numbers, if the digit at the tens place of second number is 1.

$\endgroup$
  • 2
    $\begingroup$ It took me 2 days to correctly frame this puzzle. $\endgroup$ – Nikhil Bhavar May 8 '17 at 14:21
11
$\begingroup$

This first part is the answer to the original incorrectly stated question

It seems to me only one of the multiplications is missing:

$2221065 = 3^2*5*7*11*641 = 7*11*15*1923$
$71157 = 71115+19+23$
I found this by looking at the factorisation, and seeing the large prime 641. It did not seem like the concatenation of two successive numbers in an arithmetic sequence, so then I tried combining it with a factor 3 to get 1923. This looked more promising as a concatenation of 19 and 23, and extending that sequence produced my answer.

This part is the answer to the corrected question

$11698071 = 3*7*11*89*569 = 7*11*151923$
$71157 = 71115+19+23$

This is just a variation of the previous answer.
You can however reason about it as follows. The largest number has no more than 2 digits. The sum has 5 digits so the two missing plus symbols are consecutive to make a number from three of the items. This means that the smallest item has only one digit, so it is 7 to make the first digit of the sum correct. The next largest item must be 10 or 11 because of the next two digits of the sum. So the sequence is 7/10/13/16/19 or 7/11/15/19/23. The second is then easily found to be the correct one because 71013+16+19 gives the wrong sum. Note that I didn't even need to use the product to find the answer.

$\endgroup$
  • 1
    $\begingroup$ I arrived at the same conclusion using the same method. I was trying to find a solution to the problem as stated but I think that, like you, I'm now inclined to believe the problem is incorrect. $\endgroup$ – Rubio May 8 '17 at 15:29
  • 1
    $\begingroup$ I found this solution too but since OP said he spent two days working on this problem I kept digging. It doesn't look like there are solutions for the puzzle as it is right now. $\endgroup$ – FrodCube May 8 '17 at 15:30
  • $\begingroup$ I am debuting here with this question. I am amateur, but I don't find question have any shortcoming. $\endgroup$ – Nikhil Bhavar May 8 '17 at 16:47
  • 1
    $\begingroup$ I have verified that there are no solutions with two digit numbers. $\endgroup$ – Trenin May 8 '17 at 19:10
  • 1
    $\begingroup$ @NikhilBhavar In your question you write that two additions are missing and that two multiplications are missing. In my answer there is only one multiplication missing. $\endgroup$ – Jaap Scherphuis May 9 '17 at 7:21
1
$\begingroup$

Let the numbers be $a, a+b, a+2b, a+3b$ and $a+4b$. If we assume $b=0$, then $a<11$ based on the wrong product. Even where numbers are concatenated, the concatenated number's mod 9 value is equivalent to the sum of the initial numbers, so the the sum being wrong doesn't make a difference in this aspect. The wrong sum can be written as $9k+3$, meaning the average ($a+2b$) must be written as $9k+6$. Since the digit of ones of $a+4b$ is the last ever digit in the multiplication, it must be odd, making also $a$ and $a+2b$ odd. $a$ can only be an odd digit. According to the last sentence in the question, $a+b$ must have multiple digits. 71157 must be the sum of 3 terms, so at least one of them has 5 digits. Since $a+b > (a+4b)/4$, $a+4b$ can have one digit more than $a+b$ at most. If $a+b$ has $d$ digits, $2d<6$, $d<3$, so $d$ can only be 2 and $a+b$ is a two-digit number whose digit of tens is 1. Even $a+4b$ has 2 digits. A 5-digit number can only be formed by concatenating three consecutive numbers, with the first being a single-digit number ($a$ that is).

The remaining terms in the addition are the last two numbers, whose sum is less than 199, making the $a$ in the 5-digit number 7.

$a+b$ starts with 1, which becomes the digit of thousands of the 5-digit number. That means $a+3b+a+4b = 2a+7b < 158$

That 5-digit number ends with $a+2b$, an odd 2-digit number that can be written as $9k+6$ whose digit of ones is at most 3. May be 15 or 33, but the latter means $a+b=20$, so $a+2b=15$ and $b=4$.

Solution:

The numbers are 7, 11, 15, 19 and 23. The addition is $71115+19+23=71157$ and the multiplication is $7*11*151923=11698071$.

$\endgroup$
  • $\begingroup$ Why was this downvoted despite being neither wrong nor a duplicate? Personal grudges much? $\endgroup$ – Nautilus May 10 '17 at 7:13
  • $\begingroup$ Haters gonna hate. $\endgroup$ – Nautilus May 10 '17 at 7:19
  • $\begingroup$ I haven't voted yet. $\endgroup$ – Nikhil Bhavar May 11 '17 at 3:22
  • $\begingroup$ I didn't get: "Even where numbers are concatenated, the concatenated number's mod 9 value is equivalent to the sum of the initial numbers, so the the sum being wrong doesn't make a difference in this aspect." $\endgroup$ – Nikhil Bhavar May 11 '17 at 3:24
  • $\begingroup$ Can you explain with an example. $\endgroup$ – Nikhil Bhavar May 11 '17 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.