In the product of two two-digit numbers (' '7) × (7' '), there is same digit at both the blank spaces. What can be a possible answer to this multiplication?
  $(\bf{A})\ 807\ \ \ \ (\bf{B})\ 1844\ \ \ \ (\bf{C})\ 2701\ \ \ \ (\bf{D})\ 6686$


  • 1
    $\begingroup$ Surely you must be asking for something more than to see who can work their calculators the fastest, right?? $\endgroup$ – wildBillMunson Jan 7 '17 at 18:28
  • $\begingroup$ I thought there must be a nice way to work out, but apparently not. $\endgroup$ – Sid Jan 7 '17 at 18:28
  • $\begingroup$ There is, using a quadratic expression, but in this case it takes less time to perform 8 calculator operations than it does to work out the equation. $\endgroup$ – wildBillMunson Jan 7 '17 at 18:31
  • $\begingroup$ You can derive the possible digit from the last digit in the proposed solution. For example you need 7 and 2 to make the last digit of the product 4. In the test you linked, that means that you have to calculate only 4 possible values instead of 9. (Calculators are not permitted in that test.) $\endgroup$ – M Oehm Jan 7 '17 at 18:44
  • $\begingroup$ Two-bit opinion: This is a legitimate non-calculator puzzle. Puzzles are to be solved, not hacked out with a calculator or an internet search unless they say so. $\endgroup$ – humn Jan 8 '17 at 0:02

So I finally looked at the question as printed in the test.
It's multiple choice. Often there's a way to find a pattern to save the time of brute forcing all solutions, but here the equation for what would solve these is messy:

$(10x+7)(70+x)= 10 x^2 + 707 x + 490$

This doesn't lend itself to any kind of shortcut on which digit patterns can be invalidated. The goal on these timed tests is, generally, to employ whatever strategy answers the question most quickly; since only four choices are provided, checking them isn't going to take that long in any case.

Since you know (at a minimum) that the last digit is determined exclusively by the $707x$ term and must be $7x\mod10$, you know by looking at the choices which $x$ they correspond to; work those out "brute force" (maximum of 4 multiplication problems, shouldn't take long) to find which answer it is.

We know ab initio that $x=0$ is below 800. The four listed choices in the original problem are:
  $(\bf{A})\ 807\ \ \ \ (\bf{B})\ 1844\ \ \ \ (\bf{C})\ 2701\ \ \ \ (\bf{D})\ 6686$

Quickly solve for x of 1,2,3,8.
1: 1207
2: 1944
3: 2701
8: 6786

You now know the right answer is c).

  • 1
    $\begingroup$ SO, you brute-forced each digit? Well, that works too... $\endgroup$ – Sid Jan 7 '17 at 18:27

The answer to this question can be represented by the function

f(x) = (10x + 7)(70 + x)

which can be reduced to

f(x) = 10x2 + 707x + 490.

You could generalize this to also make the other number variable with the function

f(x,y) = (10x + y)(10y + x)

which can be reduced to

f(x,y) = 10x2 + 10y2 + 101xy


If you want to be mathy about it (and I often do, but in this case this is not the best way to proceed), then start by letting the missing digit be $x$. Then the first number is $10x + 7$ and the second number is $70 + x$. Now, we want the product of these numbers to be between $800$ and $7000$. In other words, we want to know which values of $x \in \{0,1,2,3,4,5,6,7,8,9\}$ satisfy the following compound inequality: $$ 800 \le (10x+7)(70+x) \le 7000 $$ These numbers do not play nicely together, so the best way is to use a geometric approach. Graph the function $y = (10x+7)(70+x)$, focus solely on $800 \le y \le 7000$, and see which single-digit integer values of $x$ you get. You'll get $x = 1,2,3,4,5,6,7,8$.

Why did I say this isn't the best way to proceed? Because this is very time consuming (relative to brute force) and is a bad idea for what appears to be a timed test. It's much better to brute force it the way Rubio did, especially since it's a multiple choice question.


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